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1 6 . 5 6 | Geometrical Optics, , Solved Examples, JEE Main/Boards, Example 1: A person sees the point A on the rim at the, bottom of a cylindrical vessel when the vessel is empty, through the telescope T. When the vessel is completely, filled with a liquid of refractive index 1.5, he observes, a mark at the center B, of the bottom, without moving, the telescope or the vessel. What is the height of the, vessel if the diameter of its cross section is 10 cm?, Sol: The vessel is filled with a liquid of refractive index, µ when the ray from B reaches the telescope., ∴ µ sin i = 1.0 x sin r ,, , ... (i), , where ∠r =∠ADC, ∠i =∠BDC, , =, sin i, , R, =, , sin r, R 2 + h2, , Sol: For refraction of light ray from surface of sphere,, µ, µ, µ1 − µ2, ., the distance of image is obtained by 1 − 2 =, v, r, R, The ray refracted at one surface becomes object for the, opposite surface., For the first refraction surface:, , µ 1 = 1, µ 2 = µ ,u = −∞ , v = ?, r = + R, µ 1, µ −1, µR, − =, ⇒=, v, µ −1, v −∞, R, , ∴, , For refraction at the second surface:, , 2R, h2 + (2R)2, , where h is the height of the liquid, and R is the radius, of the vessel., , T, , E, , I’ serves as an object for the second surface. Now,, µ1 =µ (since light travels from sphere to air) and, µ2 =, 1,,  µR,  R (2 − µ ), u ==, BI' + , − 2R  =, µ −1,  µ −1, , , (It is positive because light travels from the left to the, right, and the distance is also in the same direction.), v = ?, r = − R ., , r, D, ,  = 1.5, , µ in the direction of one of the diameters. At what, distance from the center of the sphere, will the rays be, focused? Assume that µ < 2., , i, , R (2 − µ ), 1 µ (µ − 1) 1 − µ, − =, =, or v, v R (2 − µ), −R, 2 (µ − 1), , ∴, , h, , This distance is positive and is referred from the second, surface., , A, , B, 5 cm, , C, , ∴ Distance of the focal point from the center, , 5 cm, , =, , Substituting these values in Eq. (i),, R, µ×, =, R 2 + h2, , (, , ), , 2R, =, ⇒ µ, 2, h2 + ( 2R ), , (, , 4R 2 µ2 −=, 1 h2 4 − µ2, , =, h 10, , ), , 2 R 2 + h2, h2 + ( 2R ), , 2, , ., ,  µ2 − 1 , ⇒=, h 2R , , 2,  4 − µ , , (1.5)2 − 1, 5, = 10, = 8.45cm ., 2, 7, 4 − (1.5), , Example 2: A parallel beam of light rays is incident on, a transparent sphere of radius R and a refractive index, , R (2 − µ ), µR, ., =, +R, 2 (µ − 1), 2 (µ − 1), Q, , P, , R, A, , C, , B, , I, , I’, , Example 3: A rectangular glass block of thickness, 10 cm and refractive index 1.5 is placed over a small, coin. A beaker is filled with water of refractive index 4/3, to a height of 10 cm and is placed over the block.

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P hysi cs | 16.57, , (i) Find the apparent position of the object when it is, 1 1, −1 3, = sin−=, sin, =, 480 ., viewed from normal incidence., a, 4, µw, (ii) Draw a neat ray diagram., Obviously, therefore, TIR takes place earlier at the, (iii) If the eye is slowly moved away from the normal, at, water–air interface., a certain position, the object is found to disappear due, to the TIR. At which surface, does this happen and why?, Example 4: How long will the light take in travelling a, distance of 500 m in water? Given that µ for water is 4/3, Sol: The image of coin formed at upper surface of block,, and the velocity of light in vacuum is 3x1010 cm/s. Also, becomes object for beaker containing water. The image, calculate the equivalent path., µ1 µ2 µ1 − µ2, ., thus formed at distance v is given by, −, =, c, v, r, R, Sol: The velocity of light in a medium is given as v = ., 4, µ, For the first surface: µ=, , µ=, 1.5,, 2, 1, The optical path travelled by light is  = µ × d where d, 3, is the distance travelled in water., u=, ?, −10 cm, R =, ∞, v =, 1, , We know that, , 4, 4, ., − 1.5, 1.5, 80, 3 −, 3, =, ⇒ v1 =, − cm, v1 −10, ∞, 9, , µ=, , I’ serves as an object for the second surface. For the, second surface:, Note: This is an alternative to the apparent depth, relation., , 80 , 170, u=, cm, −BI' =, − (BA + AI') =, −  10 +, −, =, 9 , 9, , 4, µ1 = , µ2 =1, R =∞ , v 2 =?, 3, 4, 4, 1−, 1, 3 ⇒ v= 14.2cm, ∴, − 3=, 2, 170, ∞, v2, −, 9, θc: critical angle for glass–water interface, = sin−1, , 1, w, , µg, , ⇒, , θc =sin−1, , 1, w, , µa × aµg, , Velocity of light in vaccum, ., Velocity of light in water, , 4, 3x1010, ., =, 3 Velocity of light in water, , Velocity of light in water = 2.25 x 1010cm/s., Time, taken, =, , 500x100, = 2.22x10−6 sec., 2.25x1010, , Equivalent optical path, = µ x dis tance travelled in water, =, , 4, =, x500 666.64m., 3, , Example 5: In the figure shown, for an angle of, incidence i at the top surface, what is the minimum, refractive index needed for TIR at the vertical face?, , i, , Air, B, , r, , c, c, , 1, −1 8, = sin−1 .= sin, =, 62.70, 3 3, 9, x, 4 2, The critical angle for water–air interface, , Sol: Total internal reflection of light occurs inside body, when the angle of incidence is greater than critical, 1, angle θC and according to Snell’s law θC =, sin−1  , µ, Applying the Snell’s law at the top surface,

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1 6 . 5 8 | Geometrical Optics, , µ sin r =, sini ., , ... (i), , For TIR, the vertical face, , A, , µ sin θc =, 1, Using geometry, θc= 90° − r, ⇒ µ cos =, r 1 , , µ sin(90 − r)= 1, , ... (ii), , , , On squaring and adding Eqs (i) and (ii), we get, , , , ∴ µ2 sin2 r + µ2 cos2 r = 1 + sin2 i, ⇒ µ=, , , , 2, , 1 + sin i., , , , Example 6: A point source of light is placed at the, bottom of a tank containing a liquid (refractive index =, µ) up to a depth h. A bright circular spot is seen on the, surface of the liquid. Find the radius of this bright spot., Sol: The light waves incident on the surface of the water, at an incident angle greater than critical angle will get, reflected in side water. This light waves forms cone in, the volume of the tank. Thus we get relation of critical, R, angle as tan θC =, h, , c, , c, , , Rays coming out of the source and incident at an angle, greater than θc will be reflected back into the liquid;, therefore, the corresponding region on the surface will, appear dark. As it is obvious from the figure, the radius, of the bright spot is given by, , Since, =, sin θc, , cos θc, 1, ;, µ, , =, ∴R, , =i 900 − α ; =i 900 −=, α A, 0, =, β 90 −=, 2i 900 − 2A, Also,, , hsin θ, 2, , 1 − sin θc, h, 2, , µ −1, , … (i), , and =, γ 900 −=, β 2A, Thus,, , γ = r = 2A, , From geometry,, , Given µ=, g, , S, , ⇒=, R, , α 900 − A, From the figure, =, , or =, A, , 180, = 360 ., 5, , Example 8: A lens has a power of +5 dioptre in air., What will be its power if completely immersed in water?, , h, , h sin θc, , Sol: The angles of prism add up to 180o., , A+γ+, =, γ 1800., , R, , =, θc, R h tan =, , , i, i, , Sol: According to the lens maker’s formula the focal, length 1/f is, , Example 7: The cross section of the glass prism has the, form of an isosceles triangle. One of the equal faces is, coated with silver. A ray of light incident normally on, the other equal face and after getting reflected twice, emerges through the base of prism along the normal., Find the angle of the prism., ,  1, 1  µ1, 1 , =, − 1   − , , f  µ2,   R1 R 2 , , Let fa and fw be the focal lengths of the lens in air and, water, respectively, then,, , =, Pa, , ., , 3, 4, ., ; µ=, w, 2, 3, , µw, 1, =, and Pw, ., fa, fw, , =, fa 0.2m, = 20cm ., Using the lens maker’s formula,, , Pa =, , 1, = µg − 1, fa, , (, , , , , , ) R1 − R1  , , , 1, ,  1, 1  µg, 1 , =, − 1  − ,   R1 R 2 , fw  µ w, , , , 2, , , , ... (i)

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P hysi cs | 16.59, , 1  µg − µ w, = , P=, w, fw  µ w, , ,  1, 1 ,   −  …(ii),   R1 R 2 , , , Dividing Eq. (ii) by Eq. (i), we get,, , (, , ), ), , µg − µ w, Pw, 1 1, =, =, ., Pa µ µ − 1, 3 µw, w, g, ⇒, , (, ,  1   3  +5, =, Pw Pa   =, D, ,  3  4  4, , Example 9: The distance between two point sources, of light is 24 cm. Find out where you would place a, converging lens of focal length 9 cm, so that the images, of both the sources are formed at the same point., Sol: For lens the distance of the image formed from, 1 1 1, the lens is given by − = where u, v and f are, v u f, distance of object, distance of image and focal length, respectively., For S1 :, ∴, , 1, 1, 1, −, =, ν1 −x 9, , 1, 1 1, = − , ν1 9 x, , For S2 :, ∴, , 1, 1, 1, , −, =, ν2 − ( 24 − x ) 9, , Example 10: Two equiconvex lenses of focal lengths 30, cm and 70 cm, made of material of refractive index =, 1.5, are held in contact coaxially by a rubber band round, their edges. A liquid of refractive index 1.3 is introduced, in the space between the lenses filling it completely., Find the position of the image of a luminous point, object placed on the axis of the combination lens at a, distance of 90 cm from it., Sol: This system is combination of three lenses. Two, lenses of glass one lens of liquid. Add the powers to, get total power., R1 = R 2 = f1= 30cm (As µ= 1.5) ., , Similarly, R 3= R 4= f2= 70cm ., The focal length of the liquid lens (in air),, , 1, =, f3, , , , ( µ − 1)  R1, , , 1 , , R 3 , , , , , 1 , 1, =, (1.3 − 1)  −130 − 70,  = −, 70, , , , (b) m =, , … (ii), , (5.0 − 4.0 ) =, ∴m1 =, −0.25 ,, ( −4.0 ), and m2 =, , f=9 cm, S2, , 1 2, , 3 4, , ν, u, , … (i), , 1, 1, 1, ., =, −, ν2 9 24 − x, , S1, , 2, , −, , (5.0 − 1.0 ) =, ( −1.0 ), , −4.00 ., , Hence, both the images are real and inverted, the first, is magnification −0.25, and the second is −4.00., , JEE Advanced/Boards, x, , 24-x, , Since, the sign convention for S1 and S2 is just opposite., Hence,, , ν1 = −ν2 ., ⇒, , 1, 1, =, −, ν1, ν2, , ∴, , 1 1, 1, 1, −=, −, 9 x 24 − x 9, , Solving this equation, we get x = 6 cm. Therefore, the, lens should be kept at a distance of 6 cm from either, of the object., , Example 1: A 4-cm-thick layer of water covers a 6-cmthick glass slab. A coin is placed at the bottom of the, slab and is being observed from the air side along the, normal to the surface. Find the apparent position of the, coin from the surface., Air, 4cm h1, , Water, , 6cm h2, , Glass, Coin, , Sol: As the thick layer of water is placed over the, glass slab. The coin placed beneath the glass slab will

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1 6 . 6 0 | Geometrical Optics, , appear to shift upwards due to both glass and water by, distance s. This apparent shift is thus given by, , focal length for crown glass f, ω, =, = −, ', focal length for flint glass, ω', f, , 9, 325, 5, =, −, ×, =, − ;, 520, 9, 8, , , , 1 , 1 , s = s1 + s2 =  1 −  × h1 +  1 −, ×h, µ1 , µ2  2, , , , 1, f, , ', , = −, , 5, , 8f, , … (i), , The total apparent shift is, , As the focal length of the combination is 100 cm,, , , , 1 , 1 , s = h1  1 −  + h2  1 −, , µ1 , µ2 , , , , 1 1 1 1 5, 3, 1, = + = −, = =, ., F f f ' f 8f 8f 100, 3, f = × 100 =, 37.5cm ., 8, , , , 1 , 1 , s = 4 1 −,  + 6 1 −,  = 3 cm., 4, /, 3, 3, /2, , , , , f' =, , Air, 4cm h1, , h, , 6cm h2, , S, , Water, Glass, , Coin, Thus,, , h = h1 + h2 − s = 4 + 6 − 3 ., = 7.0 cm., , Example 2: An achromatic lens of focal length 100 cm, is made up of crown and flint glass lenses. Find the, focal length of each lens. Given that for crown glass, µν =, 1.5245, µr =1.5155 and for flint glass µ'ν =, 1.659, and µ'r =1.641., , −8, −8 75, × f=, × = −60 cm ., 5, 5, 2, , The achromatic doublet requires a convex lens of focal, length 37.5 cm made of crown glass and a concave lens, of focal length 60 cm made of flint glass., Example 3: A 20-cm-thick slab of glass of refractive, index 1.5 is placed in front of a plane mirror and a pin, is placed in front of it in the air at a distance of 40 cm, from the mirror. Find the position of the image., Sol: The slab of thickness t forms the image of the, object O at the point O’., The slab shifts the image of object by ( t − t µ ) . O’, serves as an object for the plane mirror., Object distance for the mirror is MO., , Sol: The focal length of the combination of lens is given, 1 1 1, by =, ., +, F f1 f2, , O, , M, , O’, , I”, , I’, , If µ and µ’ are mean refractive indices for crown and, flint glasses, respectively, then,, µν + µr 1.5245 + 1.5155, =, = 1.52 ., 2, 2, µ'ν + µr' 1.659 + 1.641, =, µ', =, = 1.65 ., 2, 2, , =, µ, , The dispersive powers ω and ω ' for crown and flint, glass, respectively, are, µν − µr 1.5245 − 1.5155 0.009, 9, =, =, ω, =, =, µ −1, 1.52 − 1, 0.52 520, , =, ω', , µ'ν − µ'r 1.659 − 1.641, 0.018, 9, =, = =, ', 1.65 − 1, 0.65, 325, µ −1, , To have an achromatic combination,, , = 40 −, , 20 100, =, cm ., 3, 3, , Since in a plane mirror, object distance = image, distance., ', =, MI' MO, =, , 100, cm ., 3, , Now, I' serves as an object for the slab again., I' is shifted to I" by 20 −, , 20 20, =cm ., 1.5, 3, , (), , ∴ Distance of the final image I" from the mirror, =, , 100 20 80, −, =, cm ., 3, 3, 3

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P hysi cs | 16.61, , Example 4: The convex surface of a thin concavo–, convex lens of glass of refractive index 1.5 has a radius, of curvature 20 cm. The concave surface has a radius, of curvature of 60 cm. The convex side is coated with, silver and placed at a horizontal surface as shown in, the figure., , For the image to be informed at the same point as the, object, u=, 2F =, 2 × 7.5 =, 15cm., The object should be placed at a distance of 15 cm, from the lens on the optical axis., (b) If fw is the focal length of lens in water, the focal, length F’ of this combination is given by, 1, ', , F, , =, , 1 1 1 1 1, + +, + +, fw fg fm fg fw, , 1, 2 2 1, =, + + , (a) Where a pin should be placed on the optical axis, ', f, fg fm, F, w, such that its image is formed at the same place?, =, fg 60, =, cm, fm 10 cm ., (b) If the concave part is filled with water of refractive, index 4/3, find the distance through which the pin, should be moved, so that the image of the pin again, coincides with pin., , Sol: When the convex side of the concavo–convex lens, is coated with silver, the combination becomes a mirror., This combination consist two lenses and one mirror, placed close to each other. The powers of all the three, will be added. When the water is filled on concave side,, we get plano-convex water lens whose focal length, is found by lens makers formula. This combination, consists of four lenses (two lenses of glass and two, lenses of water) and one mirror., (a) The refraction takes place from the first surface,, reflection from the lower surface and finally refraction, from the first surface of focal lengths fg , fm and fg ,, respectively. The combined focal length F is given by, 1 1 1 1, 2 1, = +, + = + , F fg fm fg fg fm, , … (i), , =, fm R=, 20, =, / 2 10 cm. ., 2 /2, The value of fg can be obtained by using the formula, , 1, =, fg, , (, ,  1, 1 , a, µg − 1 , −, ,  R1 R 2 , , ), ,  1, 1 , =, −, (1.5 − 1)  20, , 60 , , fg = 60 cm., , Substituting these values in Eq. (i),, 1, 2, 1, 2, =, +, =, F 60 10 15, , =, F, , 15, = 7.5cm., 2, , ... (ii), , The value of fw is calculated by using the relation,, , 1, =, fw, , (, , a, ,  1, 1 , µ w − 1 , −, ,  R1 R 2 , , ), ,  1 , 1 4, 1, =,  − 1  =, fw  3, 60, 180,  , , ⇒, , fw = 180 cm., Substituting these values in Eq. (ii), we get, 1, ', , F, , =, , 2, 2, 1, 90, ; F=, +, +, 180 60 10, 13, , ', u=, 2F, =, , 2 × 90 180, cm., =, 13, 13, , Displacement of the pin, =u − u' =15 −, , 180 15, = =1.14 cm., 13 13, , Example 5: The radius of curvature of the convex face, of plano-convex lens is 12 cm, and its µ =1.5., (a) Find the focal length of the lens., The plane surface of the lens is coated with silver., (b) At what distance from the lens, will the parallel rays, incident on the convex surface converge?, (c) Sketch the ray diagram to locate the image, when a, point object is placed on the axis at a distance of 20 cm, from the lens., (d) Calculate the image distance when the object is, placed as in (c)., Sol: Use the lens maker’s formula to find the focal, length of the plane-convex lens. When the plane side of, the lens is coated with silver, the combination becomes

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1 6 . 6 2 | Geometrical Optics, , a mirror. This combination consist two lenses and one, mirror placed close to each other. The powers of all the, three will be added., (a) The focal length f of lens of refractive index µ is, given by the formula., , 1, =, f, , , , ( µ − 1)  R1, , , 1, , −, , 1 , ., R 2 , , 1, (1.5 − 1)  12, , , , , , , , 1, 1, 1, 1, =, −, =, ν 12 20 30, , A real image is formed., , R1 = ∞ , R 2 = −12cm, µ =1.5, , , For F = 12 cm, u = +20cm and ν =?, , ν =30 cm., , In case of plano-convex lens,, , 1, ∴ =, f, , 1 1 1, + =., ν u F, , f = 24 cm., , Example 6: An object O is placed at a distance of 15, cm from a convex lens A of focal length 10 cm and its, image I1 is formed on a screen on the other side of the, lens. A concave lens B is now placed midway between A, and I1 , then the screen is moved back 10 cm to receive, a clear image I2., , (b) The focal length, fm, of plane-silvered surface is, infinity. The focal length F of the plano-convex lens,, when the plane surface is coated with silver is given by, , Find (a) The focal length of the concave lens and, , 1 1 1 1 2 1, = +, + = +, F f fm f f ∞, , Sol: The convex lens forms the image of the object at, point I1. This image acts as the object for concave lens, and the final image is formed at I1’., , ∴, , F=, , f 24, =, = 12cm., 2 2, , (b) Linear magnification of the final image., , For refraction through convex lens,, , Such a lens behaves like a concave mirror of focal, length 12 cm. The parallel rays converge at a distance, of 12 cm from the silvered surface., , u1 = OP = − 15cm ., f =+10 cm, ν1 =PI1 ., Using, , 1 1 1, − =, ν u f, A, , B, , P, , 12 cm, , 15 cm, , (c) The figure shows the ray diagram of the image, formed by this lens when the object is placed at a, distance of 20 cm from the lens. The light is incident, from the right to the left., , I1, , Q, , 15 cm, , 15 cm, , I2, , 10 cm, , 1 1 1 1  1  3−2, 1, = + =, +, =, =, ν1 f u 10  −15 , 30, 30, , ⇒ ν1 = 30 cm ., I1 serves as a virtual object for lens B. For refraction, through lens B,, , I, O, , u2 =, QI1 =, PI1 − 15 =, 30 − 15 =, +15cm,, ν2 =15 + 10 =+25cm ., 1 1 1, 1, 1, 3−5, 2, =− = −, =, =, −, f ν u 25 15, 75, 75, , (d) For a lens,, , 75, f=, −, =, −37.5cm,, 2

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P hysi cs | 16.63, , The negative sign implies it is a concave lens., Linear magnification =, , ν1, , u1, , ×, , and µ2 =− ( 0.6 + 0.4 ) =−1.0 cm, , ν2, , ( 4 / 3) −, , u2, , ν2, , 30 5, 10, = × =, −, =, −3.33, 15 3, 3, , The negative sign shows that the final image is, inverted., Example 7: A parallel beam of light travelling in water, ( µ =4 / 3) is refracted by a spherical air bubble of, radius 2 mm placed in water. Assume that the light rays, to be paraxial., (a) Find the position of the image due to refraction at, the first surface and the position of the final image., (b) Draw a ray diagram showing the positions of both, images., , ( 4 / 3) − 1, 1, =, −0.2, ( −1), , ν2 =−0.5 cm, The final image I2 is formed at a distance of 0.5 cm to, the left of the second surface P2 . The final image is at, a distance of 0.5 − 0.4 =, 0.1cm to the left of the first, surface as shown in figure., (b) The ray diagram is already shown in the figure drawn., Example 8: An object is placed at a distance of 12 cm, to the left of a diverging lens of focal length −6.0 cm. A, converging lens with a focal length of 12.0 cm is placed, at a distance d to the right of the diverging lens. Find the, distance d that corresponds to a final image at infinity., f = 6.0 cm, , f = 12 cm, , =, I2, , P1, , P2, ,  = (4/3), 0.6 cm, , 4 cm, 5 cm, , Sol: The image of the object formed by the first, refraction by the water-glass surface acts as the object, for the second refraction at glass-water surface., (a) For the refraction from a single spherical surface,, we have, µ2, ν, , −, , µ1, u, , ( µ2 − µ1 ) ., =, R, , Let P1 be pole at the first surface and P2 to be pole at, the second surface. At P,, µ1 =, ( 4 / 3) ; µ2 =1; R = +0.2cm, u1 = ∞ ., , So, , 1 ( 4 / 3) 1 − ( 4 / 3), ., −, =, ∞, +0.2, ν1, , ∴, , ν1 =−0.6 cm ., , The first surface will form a virtual image I1 at a distance, 0.6 cm to the left of P1 as shown in the figure., This image acts as an object for the second surface. So, for the second surface at P2 ,, µ1 =1,, , µ2 =( 4 / 3 ) ,, , R =−0.2cm, , 12 cm, , D, , Sol: The concave lens forms the image of the object at, point say I1. This image acts as the object for convex, lens and the final image is formed at say I1’., Applying lens formula, 1, 1, 1, −, =, ν1 −12 −6, , 1, 1, 1, −, =, ∞ ν1 − d 12, , , 1 1 1, − = twice, we have, ν u f, , ... (i), ... (ii), , Solving Eqs (i) and (ii), we have, , ν1 =− 4cm ., And d = 8cm ., Example 9: A solid glass sphere with a radius R and, a refractive index of 1.5 is coated with silver over a, hemisphere. A small object is located on the axis of the, sphere at a distance 2R to the left of the vertex of the, un-silvered hemisphere. Find the position of the final, image after all refractions and reflections have taken, place.

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1 6 . 6 4 | Geometrical Optics, , Sol: The image formed by first refraction at air-glass, spherical surface acts as the object for the concave, mirror. The image formed by the mirror acts as the, object for spherical glass-air spherical surface., The ray of light first gets refracted, then reflected and, then again refracted. For the first refraction and then, reflection, the ray of light travels from the left to the, right, while for the last refraction it travels from the, right to the left. Hence, the sign convention will change, accordingly., , Second reflection: Using, sign conventions, we have, , 1 1 1 2, with proper, + = =, ν u f R, , R, 1 1, 2, + =, − ∴ ν2 = −, 2, ν2 ∞, R, Third refraction: Again using, , µ2, ν, , −, , µ1, u, , µ2 − µ1, =, R, , with reversed sign convention, we have, , I2, I3, , O, , O, , R, , R, , 2R, First refraction: Using, , 1.5 R, µ2, , ν, sign conventions, we have, , −, , µ1, u, , µ2 − µ1, =, with proper, R, , 1.0, 1.5, 1.0 − 1.5, −, =, ν3 −1.5R, −R, , ⇒, , 1.5 1.0 1.5 − 1.0, ∴ ν1 =∞ ., −, =, ν1 −2R, +R, , R/2, , ν3 =−2R ;, , i.e. final image is formed on the vertex of the silvered, face., , JEE Main/Boards, Exercise 1, Q.1 A ray of light incident on an equilateral glass prism, shows a minimum deviation of 30°. Calculate the speed, of light through the glass prism., Q.2 Where an object should be placed from a, converging lens of focal length 20 cm so as to obtain a, real image of magnification 2?, Q.3 What changes in the focal length of (i) a concave, mirror and (ii) a convex lens occur, when the incident, violet light on them is replaced with red light?, , Q.4 State the conditions for TIR of light to take place., Calculate the speed of light in a medium, whose critical, angle is 45o., Q.5 An object is placed at the focus of concave lens., Where will its image be formed?, Q.6 Draw a graph to show the variation of the angle of, deviation ‘D’ with that of the angle of incidence ‘I’ for, a monochromatic ray of light passing through a glass, prism refracting angle ‘A’. Hence, deduce the relation., Q.7 Draw a ray diagram of an astronomical telescope, in the normal adjustment position. Write down the, expression for its magnifying power.

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P hysi cs | 16.65, , Q.8 A spherical surface of radius of curvature R separates, a rarer and a denser medium. Complete the path of the, incident ray of light, showing the formation of the real, image. Hence, derive the relation connecting an object, distance ‘u’, image distance ‘v’, radius of curvature R, and the refractive indices n1 and n2 of the two media., Briefly explain how the focal length of a convex lens, changes, with an increase in wavelength of incident, light., Q.9 (a) Draw a labelled ray diagram to show the, formation of the image by a compound microscope., Write the expression for its magnifying power., (b) How does the resolving power of a compound, microscope change and when?, (i) Refractive index of the medium between the object, and the objective lens increases and, (ii) Wavelength of the radiation used is increased., Q.10 Draw a labeled ray diagram to show the image, formation I of a refractive-type astronomical telescope., Why should the diameter of the objective of a telescope, be large?, Q.11 A beam of light converges to a point P. A lens, is placed in the path of the convergent beam at a, distance of 12 cm from P. At what point, does the beam, converge if the lens is, (i) A convex lens of focal length 20 cm., (ii) A concave lens of focal length 16cm?, Do the required calculations., Q.12 A double convex lens of glass of refractive index, 1.6 has its both surfaces of equal radii of curvature of, 30 cm each. An object of height 5 cm is placed at a, distance of 12.5 cm from the lens. Calculate the size of, the image found., Q.13 Why does the bluish color predominate in a clear, sky?, Q.14 How does the angle of minimum deviation of, a glass prism of a refractive index 1.5 change, if it is, immersed in a liquid of refractive index 1.3?, Q.15 Draw a labeled ray diagram, showing the image, formation of an astronomical telescope in the normal, adjustment position. Write the expression for its, magnifying power., , 1 1 1, Q.16 Derive the lens formula, =, − for a concave, f v u, lens, using the necessary ray diagram., , Two lenses of power 10 D and −5 D are placed in, contact., (i) Calculate the power of the new lens., (ii) Where should an object be held from the lens so as, to obtain a virtual image of magnification 2?, Q.17 Two thin lenses of power +6 D and −2 D are in, contact. What is the focal length of the combination?, Q.18 Define the refractive index of a transparent, medium. A ray of light passes through a triangular, prism. Plot a graph showing the variation of the angle, of deviation with an angle of incidence., Q.19 (a) (i) Draw a labeled ray diagram to show the, formation of image in an astronomical telescope for a, distant object., (ii) Write three distinct advantages of a reflecting-type, telescope over a refracting-type telescope., (b) A convex lens of focal length 10 cm is placed, coaxially 5 cm away from a concave lens of focal length, 10 cm. If an object is placed 30 cm in front of the convex, lens, find the position of the final image formed by the, combined system., Q.20 A converging lens is kept coaxially in contact with, a diverging lens – both the lenses being of equal focal, lengths. What is the focal length of the combination?, Q.21 (a) (i) Draw a neat labeled ray diagram of an, astronomical telescope in the normal adjustment., Explain briefly its working., (ii) An astronomical telescope uses two lenses of power, 10 D and 1 D. What is its magnifying power in the, normal adjustment?, (b) (i) Draw a neat labeled ray diagram of a compound, microscope. Explain briefly its working., (ii) Why must both the objective and the eyepiece of a, compound microscope have short focal lengths?, Q.22 Draw a labeled ray diagram of a reflecting, telescope. Mention its two advantages over the, refracting telescope.

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1 6 . 6 6 | Geometrical Optics, , Q.23 (i) An object is placed between two plane mirrors, inclined at 60o to each other. How many images do, you expect to see?, , (C) The boy cannot see his feet., , (ii) An object is placed between two plane parallel, mirrors. Why do the distant images get fainter and, fainter?, , Q.5 A point source of light S is placed in front of, two large mirrors as shown. Which of the following, observers will see only one image of S?, , (D) The boy cannot see neither his hair nor his feet., , Q.24 The magnifying power of a compound microscope, is 20. The focal length of its eye piece is 3 cm. Calculate, the magnification produced by the objective lens., Q.25 An astronomical telescope having a magnifying, power of 8 consists of two thin lenses 45 cm apart. Find, the focal length of the lenses., , B, , C, , S, 30o, , A, , (A) Only A, , (B) Only C, , Exercise 2, , (C) Both A and C, , (B) Both B and C, , Single Correct Choice Type, , Q.6 In the figure shown if the object ‘O’ moves toward, the plane mirror, then the image I (which is formed after, successive reflections from M1 & M2 , respectively) will, move:, , Q.1 Two plane mirrors are inclined at 70o . A ray incident, on one mirror at an angle of θ after reflection falls on, the second mirror and is reflected from there parallel to, the first mirror, θ is:, (A) 50, , o, , (C) 30o, , o, , (B) 45, , (D) 55o, , Q.2 There are two plane mirror with reflecting surfaces, facing each other. Both the mirrors are moving with, the speed of ν away from each other. A point object is, placed between the mirrors. The velocity of the image, formed due to the nth reflection will be, (A) nv, , (B) 2nv, , (C) 3nv, , (D) 4nv, , Q.3 A man of height ‘h’ is walking away from a street, lamp with a constant speed ‘ ν ’. The height of the street, lamp is 3h. The rate at which the length of the man’s, shadow is increasing when he is at a distance of 10h, from the base of the street lamp is:, (A) ν / 2, , (B) ν / 3, , (C) 2ν, , (D) ν / 6, , Q.4 A boy of height 1.5 m with his eye level at 1.4 m, stands before a plane mirror of length 0.75 m fixed on, the wall. The height of the lower edge of the mirror, above the floor is 0.8 m. Then,, (A) The boy will see his full image., (B) The boy cannot see his hair., , (A) Toward right, (B) Toward left, (C) With zero velocity, , O, , (D) Cannot be determined, , M2, , M1, , Q.7 A point source of light is 60 cm away from a screen, and is kept at the focus of a concave mirror that reflects, light on the screen. The focal length of the mirror is 20, cm. The ratio of average intensities of the illumination, on the screen when the mirror is present and when the, mirror is removed is:, (A) 36:1, , (B) 37:1, , (C) 49:1, , (D) 10:1, , Q.8 A concave mirror is placed on a horizontal table,, with its axis directed vertically upward. Let O be the, pole of the mirror and C its center of curvature. A point, object is placed at C. It has a real image, also located at, C (a condition called auto-collimation). If the mirror is, now filled with water, the image will be:, (A) Real and will remain at C, (B) Real and located at a point between C and ∞, (C) Virtual and located at a point between C and O, (D) Real and located at a point between C and O, Q.9 In the diagram shown below, a point source O is, placed vertically below the center of a circular plane

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P hysi cs | 16.67, , mirror. The light rays starting from the source are, reflected from the mirror such that a circular area A on, the ground receives light. Now, a glass slab is placed, between the mirror and the source O. What will be the, magnitude of the new area on the ground receiving, light?, Circular plane mirror, , Circular plane mirror, , O, , O, , Circular plane mirror, , (A) A, , Q.13 When the object is at distances u1 and u2 , the, images formed by the same lens are real and virtual,, respectively, and of same size. Then, the focal length of, the lens is:, (A), , 1, uu, 2 1 2, , (C), , u1u2, , (B), , 1, (u + u2 ), 2 1, , (D) 2 (u1 + u2 ), , Q.14 Parallel beam of light is incident on a system of, two convex lenses of focal lengths f1 = 20 cm and f2 =, 10 cm. What should be the distance between the two, lenses so that rays after refraction from both lenses get, un-deviated?, , (B) Greater than A, (C) Less than A, (D) Cannot say, as the information given is insufficient, Q.10 In the figure ABC is the cross section of a rightangled prism, and BCDE is the cross section of a glass, slab. The value of θ , so that light incident normally, on the face AB does not cross the face BC, is (given, sin−1 ( 3 / 5 ) = 37o ), , C, , (A) θ ≤ 37o, , (B) θ > 37o, , (C) θ ≤ 53o, , (D) θ < 53o, , (B) 4m, , (C) 6m, , (B) 25%, , (C) 20%, , (C) 90 cm, , (D) 40 cm, , Q.15 An object is placed at a distance of 15 cm from a, convex lens of a focal length 10 cm. On the other side, of the lens, a convex mirror is placed at its focus such, that the image formed by the combination coincides, with the object itself. The focal length of the convex, mirror is, , O, , (D) ∞, , Q.12 A point source of light is placed at a distance h, below the surface of a large deep lake. What is the, percentage of light energy that escapes directly from, the water surface? µ of the water=4/3? (Neglect partial, reflection), (A) 50%, , (B) 30 cm, , D, , Q.11 A small source of light is 4 m below the surface, of a liquid of refractive index 5/3. In order to cut off, all the light coming out of liquid surface, the minimum, diameter of the disc placed on the surface of liquid is:, (A) 3m, , (A) 60 cm, , n, , n, , , , l, , l, , A, , E, , 6/2, , 3/2, , B, , f1, , (D) 17%, , (A) 20 cm, , (B) 10 cm, , (C) 15 cm, , (D) 30 cm, , Q.16 Look at the ray diagram shown, what will be the, focal length of the 1st and the 2nd lens, if the incident, light ray is parallel to emergent ray., st, , nd, , 2, , 1, Incident, , Emergent, 5 cm, , 5 cm

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1 6 . 6 8 | Geometrical Optics, , (A) −5cm and −10cm (B) +5cm and +10cm, (C) −5cm and +5cm, , (D) +5cm and +5cm, , Q.17 A point object is kept at the first focus of a convex, lens. If the lens starts moving toward right with a, constant velocity, the image will, (A) Always move toward right, (B) Always move toward left, (C) First move toward right and then toward left., (D) First move toward left and then toward right., , (A), , π, cm, 8, , (C), , 5π, cm, 36, , π, cm, 12, π, (D) cm, 7, (B), , Q.22 Light ray is incident on a prism of angle A= 60°, and refractive index µ = 2 . The angle of incidence at, which the emergent ray grazes the surface is given by,  3 −1, 1 − 3 , (A) sin−1 , (B) sin−1 , , ,  2 ,  2 , , , , , , ,  2 , 3, (C) sin−1 ,  (D) sin−1 , ,  2 ,  3, , , , V, Object, , Previous Years’ Questions, , F, , Q.18 A ray incident at an angle 53° on a prism emerges, at an angle of 37° as shown. If the angle of incidence, is 50° , which of the following is a possible value of the, angle of emergence, , Q.1 A student measures the focal length of convex lens, by putting an object pin at a distance ‘u’ from the lens, and measuring the distance ‘v’ of the image pin. The, graph between ‘u’ and ‘v’ plotted by the student should, look like, (2002), v (cm), , v (cm), (A), , o, , 37, , (B), , o, , 53, , O, , O, , u (cm), , v (cm), , (A) 35°, , (B) 42°, , (C) 40°, , (D) 38°, , 3, and a refracting, 2, angle 90°. Find the minimum deviation produced by, the prism., , Q.19 A prism has a refractive index, , (A) 40°, , (B) 45°, , (C) 30°, , (D) 49°, , Q.20 A certain prism is found to produce a minimum, deviation of 38° . It produces a deviation of 44° when, the angle of incidence is either 42° or 62° . What is the, angle of incidence when it is undergoing a minimum, deviation?, (A) 45°, , (B) 49°, , (C) 40°, , (D) 55°, , Q.21 A thin prism of angle 5° is placed at a distance, of 10 cm from the object. What is the distance of the, image from the object? (Given µ of prism = 1.5), , u (cm), , v (cm), , (C), , (D), , O, , u (cm), , O, , u (cm), , Q.2 An experiment is performed to find the refractive, index of glass using a travelling microscope. In this, experiment, distances are measured by , (2003), (A) A vernier scale provided on the microscope, (B) A standard laboratory scale, (C) A meter scale provided on the microscope, (D) A screw gauge provided on the microscope, Q.3 Two transparent media of refractive indices, µ1 and µ3 have a solid lens shaped transparent, material of refractive index µ2 between them as shown, in the figures in Column II. A ray traversing these media, is also shown in the figures. In Column I, the different

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P hysi cs | 16.69, , relationships between µ1 , µ2and µ3 are given. Match, them to the ray diagram shown in Column II. (2007), , (A) The ray gets totally internally reflected at face CD., (B) The ray comes out through face AD., , Column I, , Column II, , (C) The angle between the incident ray and the, emergent ray is 90o., , (A) µ1 < µ2, , (p), , (D) The angle between the incident ray and the, emergent ray is 120o., , (B) µ1 > µ2, , 1, , 2, , 3, , Q.5 The focal length of a thin biconvex lens is 20cm., When an object is moved from a distance of 25 cm, in front of it to 50 cm, the magnification of its image, m, (2005), changes from m25 to m50 . The ratio 25 is, m50, , (q), , Q.6 An object at distance 2.4 m in front of a lens forms, a sharp image on a film at distance 12 cm behind the, lens. A 1-cm-thick glass plate of refractive index 1.50 is, interposed between lens and film with its plane faces, parallel to film. At what distance (from lens), should the, object be shifted to be in sharp focus on film? (2009), , 1, , 2, , 3, , (r), , (C) µ2 =µ3, , (A) 7.2 m, 3, , 1, , 2, , (C) 3.2 m, , (D) 5.6m, , Q.7 A spectrometer gives the following reading when, used to measure the angle of prism., , (s), , (D) µ, , (B) 2.4 m, , Main scale reading: 58.5°, 1, , 2, , 3, , (t), , 3, , 1, , 2, , Q.4 A ray OP of monochromatic light is incident on, the face AB of prism ABCD near vertex B at an incident, angle of 60o (see figure) If the refractive index of the, material of the prism is 3 , which of the following is, (2009), (are) correct?, B, 0, , Given that 1 division on the main scale corresponds, to 0.5°. Total divisions on the vernier scale are 30 and, match with 29 divisions of the main scale. The angle of, the prism from the above data, (2011), (A) 58.59°, , (B) 58.77°, , (C) 58.65°, , (D) 59°, , Q.8 An initially parallel cylindrical beam travels in a, medium of refractive index µ ( I ) = µ0 + µ2 I , where µ0, and µ2 are positive constants, and I is the intensity of, the light beam. The intensity of the beam is decreasing, with an increasing radius., As the beam enters the medium, it will, , (2013), , (A) Diverge, (B) Converge, (C) Diverge near the axis and converge near the, periphery, , o, , 60, P, , C, o, , 135, o, , 90, A, , Vernier scale reading: 09 divisions, , (D) Travel as a cylindrical beam 4, Q.9 The initial shape of the wave in front of the beam is, , (2000), , o, , 75, , D, , (A) Convex

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1 6 . 7 0 | Geometrical Optics, , (B) Concave, (C) Convex near the axis and concave near the periphery, (D) Planar, Q.10 An object 2.4 m in front of a lens forms a sharp, image on a film 12 cm behind the lens. A glass plate, 1cm thick, of refractive index 1.50 is interposed between, lens and film with its plane faces parallel to film. At, what distance (from lens) should object be shifted to, be in sharp focus on film? , (2012), (A) 7.2 m, , (B) 2.4 m, , (C) 3.2 m, , (D) 5.6 m, , Q.11 The graph between angle of deviation ( δ ), and angle of incidence (i) for a triangular prism is, represented by: , (2013), , (A), , , , (B), , O, , (C), , , , (D), , O, , i, , (B) 30 cm, , (C) The entire spectrum of visible light will come out of, the water at an angle of 90° to the normal., (D) The spectrum of visible light whose frequency is, less than that of green light will come out to the air, medium., Q.15 Monochromatic light is incident on a glass prism, of angle A. If the refractive index of the material of the, prism is µ , a ray, incident at an angle θ , on the face AB, would get transmitted through the face AC of the prism, provided: , (2015), , , i, , , , O, , (C) 10 cm, , A, , B, , C, , , ,  1  , (A) θ − sin−1 µ sin  A − sin−1    , ,  µ   , , i, , Q.12 Diameter of plano-convex lens is 6 cm and, thickness at the centre is 3 mm. If speed of light in, material of lens is 2 × 108 m / s , the focal length of the, lens is: , (2013), (A) 20 cm, , (B) The entire spectrum of visible light will come out of, the water at various angles to the normal., , , , O, , i, , (A) The spectrum of visible light whose frequency is, more than that of green light will come out to the air, medium., , (D) 15 cm, , , 3, Q.13 A thin convex lens made from crown glass  µ = , 2, , has focal length f. When it is measured in two different, 4, 5, and , it has the, liquids having refractive indices, 3, 3, focal lengths f1 and f2 respectively. The correct relation, between the focal lengths is: , (2014), , (A) f2 > f and f1 becomes negative, (B) f1 and f2 both become negative, (C) f1= f2 < f, (D) f1 > f and f2 becomes negative, Q.14 A green light is incident from the water to the, air – water interface at the critical angle ( θ ) . Select the, correct statement, (2014), , , ,  1  , (B) θ > cos−1 µ sin  A + sin−1    , ,  µ   , , , ,  1  , (C) θ < cos−1 µ sin  A + sin−1    ,  µ   , , , , ,  1  , (D) θ > sin−1 µ sin  A − sin−1    ,  µ   , , , Q.16 Assuming human pupil to have a radius of 0.25, cm and a comfortable viewing distance of 25 cm, the, minimum separation between two objects that human, eye can resolve at 500 nm wavelength is: , (2015), (A) 30 µm, , (B) 100 µm, , (C) 300 µm, , (D) 1 µm, , Q.17 In an experiment for determination of refractive, index of glass of a prism by i − δ , plot, it was found that, a ray incident at angle 35° , suffers a deviation of 40°, and that it emerges at angle 79° . In that case which of, the following is closest to the maximum possible value, of the refractive index ? , (2016), (A) 1.6, , (B) 1.7, , (C) 1.8, , (D) 1.5

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P hysi cs | 16.71, , JEE Advanced/Boards, , Q.1 Two flat mirrors have their reflecting surfaces, facing each other, with an edge of one mirror in contact, with an edge of the other so that the angle between, the mirrors is 60° . Find all the angular positions of, the image with respect to x-axis. Take the case when, a point object is between the mirrors at (1, 1). Point of, intersections is (0, 0) and 1st mirror is along the x-axis., , 40 cm, , Q.2 In the figure shown, AB is a plane mirror of length, 40 cm placed at a height 40 cm from ground. There, is a light source S at a point on the ground. Find the, minimum and maximum height of a man (eye height), required to see the image of the source if he is standing, at a point P on the ground as shown in the Fig., A, , 40 cm, , B, S, , P, , 20 cm 40 cm, , Q.3 A plane mirror of circular shape with radius r=20, cm is fixed to the ceiling. A bulb is placed on the axis of, the mirror. A circular area of radius R=1 on the floor is, illuminated after the reflection of light from the mirror., The height of the room is 3 m. What is the maximum, distance from the center of the mirror and the bulb so, that the required area is illuminated?, Q.4 A concave mirror of focal length 20 cm is cut into, two parts from the middle, and the two parts are moved, perpendicularly by a distance 1 cm from the previous, principal axis AB. Find the distance between the images, formed by the two parts?, , Q.5 A balloon is rising up along the axis of a concave, mirror of radius of curvature 20 m. A ball is dropped, from the balloon at a height 15 m from the mirror when, the balloon has velocity 20 m/s. Find the speed of the, image of the ball formed by a concave mirror after 4 s?, [Take: g = 10 m / s2 ], Q.6 An observer whose least distance of distinct vision, is ‘d’ views his own face in a convex mirror of radius, or curvature ‘r’. Prove that the magnification produced, r, ., cannot exceed, d + d2 + r 2, Q.7 A surveyor on one bank of canal observes the, images of the 4-inch mark and 17-ft mark on a vertical, staff, which is partially immersed in the water and held, against the bank directly opposite to him. He see that, the reflected and refracted rays come from the same, point which is the center of the canal. If the 17-ft mark, and the surveyor’s eye are both 6 ft above the water, level, estimate the width of the canal, assuming that the, refractive index of the water is 4/3. Zero mark is at the, bottom of the canal., Q.8 A ray of light travelling in air is incident at a grazing, angle (incident angle = 90° ) on a long rectangular slab, of a transparent medium of thickness t−1.0 (see figure)., The point of incidence is the origin A (0, 0). The medium, has a variable index of refraction n(y) given by, , ky 3/2 + 1, n( y ) =, , , , 1/2, , , where k =, 1.0 m−3/2 ,, , the refractive index of air is 1.0, y, P(x₁, y₁), Air, t. 1m, , Exercise 1, , M1, , (0, 0), , 10 cm, A, , 1 cm, O, , 1 cm, , M2, , B, , B(x, y), Medium x, Air, , (i) Obtain a relation between the slope (dy/dx) of the, trajectory of the ray at a point B (x, y) in the medium, and the incident angle at that point., (ii) Find the value of n sin i., (iii) Obtain an equation for the trajectory y(x) of the ray, in the medium.

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1 6 . 7 2 | Geometrical Optics, , (iv) Determine the coordinate ( x1 y1 ) of point P, where, the ray intersects the upper surface of the slab–air, boundary., , I, o, , 37, , (v) Indicate the path of the ray subsequently., , M, Q.9 A uniform, horizontal beam of light is incident, upon a quarter cylinder of radius R = 5 cm and has, a refractive index 2 / 3 . A patch on the table at a, distance ‘x’ from the cylinder is unilluminated. Find the, value of ‘x’?, , O, , 10 m, , W, , Q.14 The diagram shows five isosceles right-angled, prisms. A light ray incident at 90° at the first face, emerges at the same angle with the normal from, the last face. Find the relation between the refractive, indices?, Q.15 Two rays travelling parallel to the principal axis, , R, , 1, , X, , 2, , 3, , 4, , 5, , Q.10 An opaque cylindrical tank with an open top has, diameter of 3.00 m and is fully filled with water. When, the sunlight reaches at an angle of 37° above the, horizon, sunlight ceases to illuminate any part of the, bottom of the tank. How deep the tank is?, , strike a large plano-convex lens of a refractive index, of 1.60. If the convex face is spherical, a ray near, the edge does not pass through the focal point, (spherical aberration occurs). If this face has a radius, of curvature of 20.0 cm and the two rays are at, =, h1 0.50, =, cm and h2 12.0 cm from the principal axis,, Q.11 A beam of parallel rays of width b propagates in, then find the difference in the positions where they, a glass at an angle θ to its plane face. The beam width, cross the principal axis., after it enters into air through this face is _______ if the, refractive index of glass is µ ., , , , b, , Glass, Air, , Q.12 A parallel beam of light falls normally on the first, face of a prism of a small angle. At the second face, it is, partly transmitted and partly reflected, and the reflected, beam strikes at the first face again and emerges from, it in a direction by making an angle 6°30' with the, reversed direction of the incident beam. The refracted, beam has undergone a deviation of 1°15' from the, original direction. Find the refractive index of the glass, and the angle of the prism., Q.13 A light ray I is incident on a plane mirror M. The, mirror is rotated in the direction as shown in the figure, by an arrow at a frequency 9 / π rev / sec . The light, reflected by the mirror is received on the wall W at a, distance 10 m from the axis of rotation. When the angle, of incidence becomes 37o, find the speed of the spot (a, point) on the wall?, , C, , R, , x, , Q.16 A room contains air in which the speed of sound, is 340m/s. The walls of the room are made of concrete,, in which the speed of sound is 1700m/s. (a) Find the, critical angle for the TIR of sound at the concrete–air, boundary. (b) In which medium must the sound be, undergone the TIR?, Q.17 A rod made of glass ( µ =1.5 ) and of square cross, section is bent as shown in the figure. A parallel beam, of light falls perpendicularly on the plane flat surface A., Referring to the diagram, d is the width of a side, and, R is the radius of inner semicircle. Find the maximum, d, value of ratio, so that all the light rays entering the, R, glass through surface A emerge from the glass through, surface B.

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P hysi cs | 16.73, , of the bulb. (c) At this 7.50-m distance, a convex lens is, set up with its axis pointing toward the bulb. The lens, has a circular face with a diameter of 15.0 cm and a, focal length of 30.0 cm. Find the diameter of the image, of the bulb formed on a screen kept at the location of, the image. (d) Find the light intensity at the image., , d, A, , 3, , Q.23 A thin plano-convex lens fits exactly into a planoconcave lens with their plane surface parallel to each, other as shown in the figure. The radius of curvature, Q.18 A prism of refractive index 2 has a refracting, of the curved surface R = 30 cm. The lenses are, angle of 30° . One of the refracting surfaces of the, made of different material having a refractive index, prism is polished. For the beam of monochromatic light =, µ1 3 / 2and, =, µ2 5 / 4 as shown in the Fig., to retrace its path, find the angle of incidence on the, refracting surface., Q.19 An equilateral prism deviates a ray by 23° for, two angles of incidence differing by 23° . Find µ of the, prism?, Q.20 A ray is incident on a glass sphere as shown, in the figure The opposite surface of the sphere is, partially coated with silver. If the net deviation of the, ray transmitted at the partially silvered surface is 1/3rd, of the net deviation suffered by the ray reflected at the, partially silvered surface (after emerging out of the, sphere), find the refractive index of the sphere., , o, , 60, , Partially, silvered, , Q.21 Two thin similar watch glass pieces are joined, together front to front, with rear convex portion is, coated with silver, and the combination of glass pieces, is placed at a distance α =60 cm from a screen. A, small object is placed normal to the optical axis of the, combination such that its two times magnified image, is formed on the screen. If air between the glass pieces, is replaced by water ( µ =4 / 3) , calculate the distance, through which the object must be displaced so that a, sharp image is again formed on the screen., Q.22 A spherical light bulb with a diameter of 3.0 cm, radiates light equally in all directions, with a power of, 4.5π W. (a) Find the light intensity at the surface of the, bulb. (b) Find the light intensity 7.50 m from the center, , 1=3/2, , 1=5/4, , (i) If plane surface of the plano-convex lens is coated, with silver, then calculate the equivalent focal length, of this system and also the nature of this equivalent, mirror., (ii) An object having a transverse length of 5 cm in, placed on the axis of equivalent mirror (in par 1) at, a distance 15 cm from the equivalent mirror along, the principal axis. Find the transverse magnification, produced by the equivalent mirror., Q.24 Two identical convex lenses L1 and L2 are placed, at a distance of 20 cm from each other on the common, principal axis. The focal length of each lens is 15 cm,, and lens L2 is placed right to lens L1. A point object, is placed at a distance of 20 cm left to lens L1 on, the common principal axis of two lenses. Find where, a convex mirror of radius of curvature 5 cm should, be placed to the right of L2 so that the final image, coincides with the object?, Q.25 A thin equiconvex lens of glass of a refractive, index µ =3 / 2 and a focal length 0.3 m in air is sealed, into an opening at one end of a tank filled with water, ( µ =4 / 3) . On the opposite side of the lens, a mirror, is placed inside the tank on the tank wall perpendicular, to the lens axis, as shown in the figure. The distance, between the lens and the mirror is 0.8 m. A small object, is placed outside the tank in front of the lens at a, distance of 0.9 m from the lens along its axis. Find the

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1 6 . 7 4 | Geometrical Optics, , position (relative to the lens) of the image of the object, formed by the system., , (B) f, I/4, , (C) 3f/4, I/2, , (D) f, 3I/4, , 0.8 m, , Exercise 2, Single Correct Choice Type, Q.1 An object is placed in front of a convex mirror, at a distance of 50 cm. A plane mirror is introduced, covering the lower half of the convex mirror. If the, distance between the object and the plane mirror is 30, cm, there is no gap between the images formed by the, two mirrors. The radius of the convex mirror is:, (A) 12.5 cm, , (B) 25 cm, , (C) 50 cm, , (D) 100 cm, , Q.2 An infinitely long rod lies along the axis of a, concave mirror of a focal length f. the near end of the, rod is at a distance u>f from the mirror. Its image will, have a length., (A) f 2 / (u − f ), , (B) uf / (u − f), , (C) f 2 / (u + f ), , (D) uf / (u + f ), , Q.3 A luminous point object is moving along the, principal axis of a concave mirror of a focal length 12, cm toward it. When its distance from mirror is 20 cm,, its velocity is 4 cm/s. The velocity of the image in cm/s, at that instant is:, (A) 6 toward the mirror, (B) 6 away from the mirror, (C) 9 away from the mirror, (D) 9 toward the mirror, Q.4 A thin lens has a focal length f, and its aperture, has a diameter d. It forms an image of intensity I. Now, the central part of the aperture up to diameter (d/2), is blocked by an opaque paper. The focal length and, image intensity would change to, , Q.5 An object is placed in front of a thin convex lens, of focal length 30 cm, and a plane mirror is placed 15, cm behind the lens. If the final image of the object, coincides with the object, the distance of the object, from the lens is, (A) 60 cm, , (B) 30 cm, , (C) 15 cm, , (D) 25 cm, , Q.6 A converging lens of a focal length 20 cm and, diameter 5 cm is cut along line AB. The part of the lens, shaded in the diagram is used to form an image of a, point P placed 30 cm away from it on line XY, which is, perpendicular to the plane of the lens. The image of P, will be formed., , 2 cm, , 2 cm, , 5 cm, , Mirror, , 0.9 m, , (A) f/2, I/2, , A, , B, Y, , X P, 30 cm, , (A) 0.5 cm above XY, , (B) 1 cm below XY, , (C) On XY (D) 1.5 cm below XY, Q.7 A screen is placed 90 cm away from an object. The, image of the object on the screen is formed by a convex, lens at two different locations separated by 20 cm. The, focal length of the lens is, (A) 18 cm, , (B) 21.4 cm, , (C) 60 cm, , (D) 85.6 cm, , Q.8 In the above problem, if the sizes of the images, formed on the screen are 6 cm and 3 cm, then the, height of the object is nearly:, (A) 4.2 cm (B) 4.5 cm, (C) 5 cm (D) 9 cm, Q.9 A concave mirror cannot form, (A) A virtual image of a virtual object, (B) A virtual image of a real object, (C) A real image of a real object, (D) A real image of a virtual object

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P hysi cs | 16.75, , Multiple Correct Choice Type, Q.10 A reflecting surface is represented by the equation,  πx , 2L, =, y, sin   ,0 ≤ × ≤ L . A ray travelling horizontally, π,  L , becomes vertical after reflection. The coordinates of, the point(s) where this ray is incident is, , y, , 1, , 2, , 3, (A) µ1 < µ2 < µ3, , (B) µ12 − µ22 > µ32, , (C) µ12 − µ32 > µ22, , (D) µ12 + µ22 > µ32, , Q.13 For refraction through a small angle prism, the, angle of deviation:, (A) Increases with an increase in RI of prism., (B) Will decrease with an increase in RI of prism.,  L 2L , (A)  ,, , 4 π , , , ,  L 3L , (B)  ,, , 3 π , , , ,  3L 2L , (C)  ,, ,  4 π , , , ,  2L 3L , (D)  ,, ,  3 π , , , , (C) Is directly proportional to the angle of prism., (D) Will be 2 D for a ray of RI = 2.4 if it is d for a ray of, RI = 1.2., Q.14 For the refraction of light through a prism, , Q.11 In the figure shown, consider the first reflection at, the plane mirror and second at the convex mirror. AB, is object., , V, A, 10cm, , (A) For every angle of deviation, there are two angles, of incidence., (B) The light travelling inside an equilateral prism is, necessarily parallel to the base when prism is set for a, minimum deviation., (C) There are two angles of incidence for a maximum, deviation., , B, , C, 10cm, , 120cm, , 50cm, (A) The second image is real, inverted by 1/5th, magnification w.r.t. AB., (B) The second image is virtual and erect by 1/5th, magnification w.r.t. AB., (C) The second image moves toward the convex mirror., (D) The second image moves away from the convex, mirror., Q.12 In the diagram shown, a ray of light is incident, on the interface between 1 and 2 at an angle slightly, greater than the critical angle. The light undergoes, TIR at this interface. After that, the light ray falls at, interfaces 1 and 3, and again it undergoes TIR. Which, of the following relations should hold true?, , (D) The angle of minimum deviation will increase if, refractive index of prism is increased keeping the, outside medium unchanged if µp > µs ., Q.15 A man of height 170 cm wants to see his complete, image in a plane mirror (while standing). His eyes are at, a height of 160 cm from the ground., (A) Minimum length of the mirror=80 cm., (B) Minimum length of the mirror=85 cm., (C) Bottom of the mirror should be at a height 80 cm, or less., (D) Bottom of the mirror should be at a height 85 cm., Q.16 A flat mirror M is arranged parallel to a wall W, at a distance l from it. The light produced by a point, sources S kept on the wall is reflected by the mirror and, produces a light spot on the wall. The mirror moves, with a velocity v toward the wall.

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1 6 . 7 6 | Geometrical Optics, , Wall, , O, , W, , S, , B, , l, , A, , V, I, , M, , (A) The spot of light will move with the speed v on the, wall., (B) The spot of light will not move on the wall., (C) As the mirror comes closer, the spot of light will, become larger and shift away from the wall with a, speed larger than v., (D) The size of the light spot on the wall remains the, same., Q.17 Two reflecting media are separated by a spherical, interface as shown in the figure. PP’ is the principal axis;, µ1 and µ2 are the medium of refraction, respectively,, then,, , 2, P, , (A) x = 40 cm, , (B) y =+1 cm, , (C) z = +1 cm, , (D) z=−1 cm, , Q.20 Which of the following can form a diminished,, virtual and erect image of your face., (A) Converging mirror (B) Diverging mirror, (C) Converging lens, , (D) Diverging lens, , Assertion Reasoning Type, , 1, P, , (A) If µ2 > µ1 , then there cannot be a real image of a, real project., (B) If µ2 > µ1 , then there cannot be a real image of a, virtual object., (C) If µ1 > µ2 , then there cannot be a virtual image of a, virtual object., (D) If µ1 > µ2 , then there cannot be a real image of a, real object., Q.18 A luminous point object is placed at O. whose, image is formed at I as shown in the figure. AB is the, optical axis. Which of the following statements are, correct?, (A) If a lens is used to obtain an image, the lens must, be converging., (B) If a mirror is used to obtain an image, the mirror, must be a convex mirror having a pole at the point of, intersection of lines OI and AB., (C) Position of the principal focus of mirror cannot be, found., (D) I is a real image., , Q.19 A lens is placed in the XYZ coordinate system such, that its optical center is the origin and the principal axis, is along the x-axis. The focal length of the lens is 20 cm., A point object has been placed at the point (-40 cm, +1, cm, −1 cm). Which of the following are correct about, coordinates of the image?, , Q.21 Statement-I: If a source of light is placed in front, of rough wall, its image is not seen., Statement-II: The wall does not reflect light., (A) Statement-I is true, and statement-II is true;, statement-II is correct explanation for statement-I, (B) Statement-I is true, and statement-II is true; statement, -II is NOT the correct explanation for statement-I., (C) Statement-I is true, and statement-II is false., (D) Statement-I is false, and statement-II is true., Q.22 Statement-I: As the distance x of a parallel ray, from axis increases, the focal length decreases, , X, P, , A, , C Axis, , Statement-II: As x increases, the distance from the, pole to the point of intersection of a reflected ray with, the principal axis decreases., (A) Statement-I is true, and statement-II is true;, statement-II is correct explanation for statement-I.

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P hysi cs | 16.77, , (B) Statement-I is true, and statement-II is true;, statement-II is NOT the correct explanation for, statement-I., (C) Statement-I is true, and statement-II is false., (D) Statement-I is false, and statement-II is true., , (A) Statement-I is true, statement-II is true, and, statement-II is the correct explanation for statement-I., (B) Statement-I is true, statement-II is true, and, statement-II is NOT the correct explanation for, statement-I., (C) Statement-I is true, and statement-II is false., , Q.23 Statement-I: When an object dipped in a liquid is, viewed normally, the distance between the image and, the object is independent of the height of the liquid, above the object., , (D) Statement-I is false, and statement-II is true., , Statement-II: The normal shift is independent of, the location of the slab between the object and the, observer., , Paragraph 1: Spherical aberration in spherical mirrors, is a defect that is due to the dependence of focal length, ‘f’ on the angle of incidence ' θ ' as shown in the figure, is given by, , (A) Statement-I is true, statement-II is true, and, statement-II is the correct explanation for statement-I., (B) Statement-I is true, statement-II is true, and, statement-II is NOT the correct explanation for, statement-I., , Comprehension Type, , f=, R−, , R, sec θ ,, 2, , (D) Statement-I is false, and statement-II is true., , where R is radius of curvature of mirror and q is the angle, of incidence. The rays that are close to the principal axis, are called marginal rays. As a result, different rays focus, at different points and the image of a point object is, not a point., , Q.24 Statement-I: When two plane mirrors are kept, perpendicular to each other as shown in the figure (O is, the point object), three images will be formed., , Q.26 If fp and fm represent the focal length of paraxial, and marginal rays, respectively, then the correct, relationship is:, , (C) Statement-I is true, and statement-II is false., , O, , Statement-II: In case of a multiple reflection, the image, of one surface can act as an object for the next surface., (A) Statement-I is true, statement-II is true, and, statement-II is the correct explanation for statement-I., (B) Statement-I is true, statement-II is true, and, statement-II is NOT the correct explanation for, statement-I., (C) Statement-I is true, and statement-II is false., (D) Statement-I is false, and statement-II is true., Q.25 Statement-I: Keeping a point object fixed, if a, plane mirror is moved, the image will definitely move., Statement-II: In case of a plane mirror, the distance, between a point object and its image from a given, point on mirror is equal., , (A) fp = fm, , (B) fp > fm, , (C) fp < fm, , (D) none, , Q.27 If the angle of incidence is 60o, then the focal, length of this ray is:, (A) R, , (B) R/2, , (C) 2R, , (D) 0, , Paragraph 2: A student is performing Young’s double, slit experiment. There are two slits S1 and S2. The, distance between them is d. There is a large screen, at a distance D (D >>d) from the slits. The setup is, shown in the following figure. A parallel beam of light, is incident on it. A monochromatic light of wavelength, λ is used. The initial phase difference between the two, slits behave as two coherent sources of light is zero., The intensities of light waves on the screen coming, out of S1 and S2 are same, i.e. I0 . In this situation, the, principal maximum is formed at point P. At the point, on screen where the principal maximum is formed, the, phase difference between two interfering waves is zero.

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1 6 . 7 8 | Geometrical Optics, , point P (see figure). The value of α is, , Screen, S, O d, S, , (A) sin−1, , P, (C) sin−1, , t (µ − 1), d, t (µ − 1)D, , (A) 180o , , (B) 90o, , (C) Cannot be determined (D) None, Q.29 For paraxial rays, focal length approximately is, (A) R, , (B) R/2, , (C) 2R, , (D) none, , Q.30 Which of the following statements are correct, regarding spherical aberration:, (A) It can be completely eliminated., (B) It cannot be completely eliminated, but it can be, minimized by allowing either paraxial or marginal rays, to hit the mirror., (C) It is reduced by taking mirrors with large aperture., (D) None of these., Q.31 Initially, the distance of third minima from principal, maxima will be, (A), , 3λD, 2d, , (B), , 3λD, d, , (C), , 5λD, 4d, , (D), , 5λD, 2d, , Q.32 A glass slab of thickness t and refractive index, µ is introduced before S2 . Now, P does not remain, the point of principal maximum. Suppose the principal, maximum forms at a point P’ on screen, then PP’ is, equal to, (A), (C), , tD ( µ − 1 ), d, D (µ − 1), t, , (B), (D), , tD ( µ − 1 ), , d, , Q.33 Use the statement given in previous question., Now, a parallel beam is incident at an angle α w.r.t. line, OP, such that the principal maximum again comes at, , t (µ − 1), d, tD, d, , Q.34 Match the Column, Column I, , Column II, , (A) Conversing system, , (p) Convex lens, , (B) Concave lens, , (q) Concave lens, , (C) A virtual image is formed by, , (r) Concave mirror, , (D) Magnification < 1 is possible, with, , (s) Convex mirror, , Previous Years’ Questions, Q.1 A student performed the experiment of, determination of the focal length of a concave mirror, by u-v method using an optical bench of length 1.5, m. The focal length of the mirror used is 24 cm. The, maximum error in the location of the image can be 0.2, cm. The five sets of (u, v) values recorded by the student, (in cm) are: (42, 56), (48, 48), (60, 40), (66, 33) and (78,, 39). The data set (s) that cannot come from experiment, and is (are) incorrectly recorded, is (are), (1999), (A) (42, 56), , (B) (48, 48), , (C) (66, 33), , (D) (78, 39), , Q.2 A light beam travels from Region I to Region IV, (See figure). The refractive index in Regions I, II, III and, n, n, IV are n0 , 0 and 0 , respectively. The angle of incidence, 6, 8, θ for which the beam misses entering Region IV (as in, the figure):, (2004), Region I, n0, , , , 2d, D (µ − 1), , (D) sin−1, , d, , D, Q.28 The total deviation suffered by the ray falling on, the mirror at an angle of incidence 60o is, , (B) cos−1, , 0, , Region II, , Region III, , Region IV, , n0, 2, , n0, 6, , n0, 8, , 0.2 m, , 3, (A) sin−1  , 4, , 1, (B) sin−1  , 8, , 1, (C) sin−1  , 4, , 1, (D) sin−1  , 3, , 0.6 m

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P hysi cs | 16.79, , Q.3 An optical component and an object s placed, along its optical axis are given in column I. The distance, between the object and the component can be varied., The properties of images are given in column II. Match, all the properties of images from column II with the, appropriate components given in column I. Indicate, your answer by darkening the appropriate bubbles of, the 4 x 4 matrix given in the ORS., (2006), , (C) The intensity of the characteristics X-rays depends, on the electrical power given to the X-ray tube., (D) Cutoff wavelength of the continuous X-rays depends, on the energy of the electrons in the X-ray tube., , Air, , 1, , (A), , Column I, , (B), Meta-material, , Column II, , 1, , Air, , Meta-material, 2, , 2, , (A), , S, , (p) Real image, , (B), , Air, , S, , (q) Virtual image, , (C), , S, , (D), , S, , (r) Magnified image, , (s) Image at infinity, , Q.4 Two beams of red and violet colors are pass, separately through a prism (angle of the prism is 60o)., In the position of minimum deviation, the angle of, refraction will be, (2007), (A) 30o for both the colors, , (B) Greater for the violet color, (C) Greater for the red color, , (D) Equal but not 30o for both the colors, Q.5 Which one of the following statements is WRONG, in the context of X-rays generated from an X-ray tube?, , (2001), (A) The wavelength of the characteristics X-rays, decreases when the atomic number of the target, increases., (B) The cutoff wavelength of the continuous X-rays, depends on the atomic number of the target., , 1, , (C), Meta-material, , Air, (D), , 2, , 1, , Meta-material, , 2, , Paragraph: Most materials have a refractive index n >1., Therefore, when a light ray from air enters a naturally, sin θ1 n2, =, occurring material, then by the Snell’s law,, sin θ2 n2, , it is understood that the refracted ray bends toward, the normal, but it never emerges on the same side, of the normal as the incident ray. According to the, electromagnetism, the refractive index of the medium is, c, given by the relation, n =   = ± ε1µ1 , where c is the, v, speed of electromagnetic waves in vacuum, v is its speed, in the medium, ε1 and µ1 are the relative permittivity, and permeability of the medium, respectively. In a, normal material, both ε1 and µ1 are positive, implying, positive n for the medium. When both ε1 and µ1 are, negative, one must choose the negative root of n. Such, materials with negative refractive indices can now be, artificially prepared and are called meta-materials., They exhibit a significantly different optical behavior,, without violating any physical laws. Since n is negative,, it results in a change in the direction of propagation, of the refracted light. However, similar to the normal, materials, the frequency of light remains unchanged, upon refraction even in meta-materials., (2012)

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1 6 . 8 0 | Geometrical Optics, , Q.6 For light incident from air on a meta-material, the, appropriate ray diagram is , , (B) Real and at a distance of 16 cm from the mirror, , (C) Virtual and at a distance of 20 cm from the mirror, (D) Real and at a distance of 20 cm from the mirror, , 1, (A), , 1, (B), , Air, Meta-material, , Air, Meta-material, , 2, , 2, , 1, (C), , 1, (D), , Air, Meta-material, , Air, Meta-material, , 2, , 2, , Q.10 The image of an object approaching a convex, mirror of a radius of curvature 20 m along its optical, 25, 50, axis moves from, m to m in 30 s. What is the, 3, 7, speed of the object in km/h?, (2011), Q.11 A bi-convex lens is formed with two thin planoconvex lenses as shown in the figure. Refractive index n of, the first lens is 1.5 and that of the second lens is 1.2. Both, the curved surfaces are of the same radius of curvature R, = 14 cm. For this bi-convex lens, for an object distance of, 40 cm, the image distance will be – , (2012), n = 1.5, , Q.7 Choose the correct statement., , n = 1.2, , (A) The speed of light in the meta-material is v= c n ., (B) The speed of light in the meta-material is v=, , c, ., n, , (C) The speed of light in the meta-materials is v = c., (D) The wavelength of the light in the meta-material, , ( λm ), , is given by λm =λair n , where λair is the, , wavelength of the light in air., Q.8 A biconvex lens is formed with two thin planoconvex lenses as shown in the figure. Refractive index n, of the first lens is 1.5 and that of the second lens is 1.2., Both curved surfaces are of same radius of curvature, R = 14 cm. For this biconvex lens, for an object distance, of 40 cm, the image distance will be, (2009), n = 1.5, , R = 14 cm, , (A) – 280.0 cm, , (B) 40.0 cm, , (C) 21.5 cm, , (D) 13.3 cm, , Q.12 A transparent slab of thickness d has a refractive, index n (z) that increases with z. Here z is the vertical, distance inside the slab, measured from the top. The slab, is placed between two media with uniform refractive, indices n1 and n2 ( > n1 ) , as shown in the figure. A ray, of light is incident with angle θ1 from medium 1 and, emerges in medium 2 with refraction angle θf with a, lateral displacement l., (2016), Which of the following statement (s) is (are) true?, , n = 1.2, , 1, n1 = constant, , 1, , n(z), z, , d, , R = 14 cm, , Q.9 A biconvex lens of focal length 15 cm is in front of, a plane mirror. The distance between the lens and the, mirror is 10 cm. A small object is kept at a distance of, 30 cm from the lens. The final image is , (2011), (A) Virtual and at a distance of 16 cm from the mirror, , n2 = constant, , l, , 2

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P hysi cs | 16.81, , =, θi n2 sin θf, (A) n1 sin, (B) n1 sin θ=, i, , (n2 − n1 ) sin θf, , (C) l is independent of n2, (D) l is dependent on n (z), Q.13 A small object is placed 50 cm to the left of a thin, convex lens of focal length 30 cm. A convex spherical, mirror of radius of curvature 100 cm is placed to the, right of the lens at a distance of 50 cm. The mirror is, tilted such that the axis of the mirror is at an angle, θ= 30° to the axis of the lens, as shown in the figure., , (2015), , Q.16 A right angled prism of refractive index µ1, is, placed in a rectangular block of refractive index µ2,, which is surrounded by a medium of refractive index, µ3 , as shown in the figure. A ray of light ‘e’ enters the, rectangular block at normal incidence. Depending, upon the relationships between µ1 , µ2 and µ3 , it takes, one of the four possible paths ‘ef’, ‘eg’, ‘eh’, or ‘ei’., f, o, , 45, , e, , 1, h, , i, , f = 30 cm, , 2, , , (-50, 0), , x, , (0, 0), , R = 100 cm, 50 cm, , List I, , If the origin of the coordinate system is taken to be at, the centre of the lens, the coordinates (in cm) of the, point (x, y) at which the image is formed are (2016), , (, ) , (C) (50 − 25 3 , 25 ), , (A) 25, 25 3, , (, , (B) 125 / 3, 25 / 3, , ), , P., Q., R., , (D) (0, 0), , S., , (, , ), , 1 ˆ, i + 3 ˆj, Q.14 A ray of light travelling in the direction, 2, is incident on a plane mirror. After reflection, it travels, , is, (A) 30°, , (, , ), , 1 ˆ, i + 3 ˆj . The angle of incidence, 2, (2013), , (B) 45°, , (C) 60°, , (D) 75°, , Q.15 The image of an object, formed by a plano-convex, lens at a distance of 8 m behind the lens, is real and is, one-third the size of the object. The wavelength of light, 2, inside the lens is times the wavelength in free space., 3, The radius of the curved surface of the lens is  (2013), (A) 1 m, , (B) 2 m, , (C) 3 m, , (D) 4 m, , 3, , Match the paths in list I with conditions of refractive, indices in list II and select the correct answer using the, codes given below the lists:, (2013), , (50+503, -50), , along the direction, , g, , List II, , e→ f, e→ g, , e→ h, e→ i, , 1., , µ1 > 2 µ2, , 2., , µ2 > µ1 and µ2 > µ3, , 3., , µ1 = µ2, , 4., , µ2 < µ1 < 2 µ2 and, , µ 2 > µ3, , Codes:, P, , Q, , R, , S, , A, , 2, , 3, , 1, , 4, , B, , 1, , 2, , 4, , 3, , C, , 4, , 1, , 2, , 3, , D, , 2, , 3, , 4, , 1, , Q.17 A transparent thin film of uniform thickness, and refractive index n1 = 1.4 is coated on the convex, spherical surface of radius R at one end of a long solid, glass cylinder of refractive index n2 = 1.5 , as shown, in the figure. Rays of light parallel to the axis of the, cylinder traversing through the film from air to glass

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1 6 . 8 2 | Geometrical Optics, , get focused at distance f1 from the film, while rays of, light traversing from glass to air get focused at distance, f2 from the film. Then, (2014), , List I, , List II, , R, , -r, , S, , r, , n1, , n2, , Air, , (A) f1 = 3R, , (B) f1 = 2.8R, , (C) f2 = 2R, , (D) f2 = 1.4R, Code:, , Q.18 A point source S is placed at the bottom of a, transparent block of height 10 mm and refractive index, 2.72. It is immersed in a lower refractive index liquid as, shown in the figure. It is found that the light emerging, from the block to the liquid forms a circular bright spot, of diameter 11.54 mm on the top of the block. The, refractive index of the liquid is , (2014), , Liquid, Block, (A) 1.21, , (B) 1.30, , S, , (C) 1.36, , (D) 1.42, , Q.19 Four combinations of two thin lenses are given in, list I. The radius of curvature of all curved surfaces is r, and the refractive index of all the lenses is 1.5. Match, lens combinations in List I with their focal length in, list II and select the correct answer using the code given, below the lists. , (2014), List I, , List II, , P, , 2r, , Q, , r/2, , (A) P-1, Q-2, R-3, S-4, (B) P-2, Q-4, R-3, S-1, (C) P-4, Q-1,R-2, S-3, (D) P-2, Q-1, R-3, S-4, Q.20 Consider a concave mirror and a convex lens, (refractive index = 1.5) of focal length 10 cm each,, separated by a distance of 50 cm in air (refractive, index = 1) as shown in the figure. An object is placed, at a distance of 15 cm from the mirror. Its erect image, formed by this combination has magnification M1 ., When the set-up is kept in a medium of refractive index, 7/6, the magnification becomes M2 . The magnitude, , M2, M1, , is, , (2015), , Q.21 Two identical glass rods S1 and S2 (refractive, index = 1.5) have one convex end of radius of curvature, 10 cm. They are placed with the curved surfaces at a, distance d as shown in the figure, with their axes (shown, by the dashed line) aligned. When a point source of, light P is placed inside rod S1 on its axis at a distance of, 50 cm from the curved face, the light rays emanating, from it are found to be parallel to the axis inside S2 ., The distance d is , (2015), , S1, , S2, , P, 50 cm, , d, , (A) 60 cm, , (B) 70 cm, , (C) 80 cm, , (D) 90 cm

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P hysi cs | 16.83, , Q.22 A monochromatic beam of light is incident at, 60° on one face of an equilateral prism of refractive, index n and emerges from the opposite face making an, , angle θ (n) with the normal (see the figure). For n = 3, dθ, the value of θ is 60° and, = m. The value of m is, dn, , (2015), , (D) NA of S1 placed in air is the same as that of S2 placed, in water., Q.24 A parallel beam of light is incident from air at an, angle α on the side PQ of a right angled triangular, prism of refractive index n = 2 . Light undergoes total, internal reflection in the prism at the face PR when α, has a minimum value of 45° . The angle θ of the prism, is , (2016), , , , o, , 60, , (C) NA of S1 placed in air is the same as that of S2, 4, immersed in liquid of refractive index, ., 15, , P, , , Paragraph 1: Light guidance in an optical fiber can, be understood by considering a structure comprising, of thin solid glass cylinder of refractive index n1, surrounded by a medium of lower refractive index n2., The light guidance in the structure takes place due to, successive total internal reflections at the interface of, the media n1 and n2 as shown in the figure. All rays, with the angle of incidence i less than a particular value, im are confined in the medium of refractive index n1., The numerical aperture (NA) of the structure is defined, as sin im, n1 > n2, Air, , Cladding, , , i, , n2, , Core, n1, , , , n =2, R, , Q, , (A) 15°, , (B) 22.5°, , (C) 30°, , (D) 45°, , Q.25 A plano-convex lens is made of a material of, refractive index n. When a small object is placed 30, cm away in front of the curved surface of the lens, an, image of double the size of the object is produced., Due to reflection from the convex surface of the lens,, another faint image is observed at a distance of 10 cm, away from the lens. Which of the following statement(s), is(are) true? , (2016), (A) The refractive index of the lens is 2.5, (B) The radius of curvature of the convex surface is, 45 cm, (C) The faint image is erect and real, , Q.23 For two structures namely S1 with n1 = 45 / 4, and n2 = 3 / 2 and S2 with n1 = 8 / 5 and n2 = 7 / 5, and taking the refractive index of water to be 4/3 and, that of air to be 1, the correct option(s) is(are)  (2015), (A) NA of S1 immersed in water is the same as that of, 16, S2 immersed in a liquid of refractive index, 3 15, 6, (B) NA of S1 immersed in liquid of refractive index, 15, is the same as that of S2 immersed in water, , (D) The focal length of the lens is 20 cm

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1 6 . 8 4 | Geometrical Optics, , MASTERJEE Essential Questions, JEE Main/Boards, , JEE Advanced/Boards, , Exercise 1, , Exercise 1, , Q.11, , Q.12, , Q.24, , Exercise 2, , Q.3, , Q.4, , Q.8, , Q.13, , Q.22, , Q.23, , Exercise 2, , Q. 3, , Q.5, , Q.7, , Q.6, , Q.9, , Q.11, , Q.8, , Q.10, , Q.18, , Q.24, , Q.25, , Q.30, , Q.23, , Q.27, , Q.37, , Q.31, , Q.33, , Answer Key, JEE Main/ Boards, Exercise 1, Q.1 2.12 × 108 ms−1, 3 × 10, , Q.2 u=−30 cm, , 8, , Q.5 The image is formed on the same side of object., m / s , 2, Q.9 (b) (i) When the refractive index of the medium increases, the resolving power increases., Q.4, , (ii) When the wavelength of the radiation increases, the resolving power decreases., Q.11 For convex lens +7.5 cm, for concave lens +48 cm., Q.12 A virtual image of height 10 cm is formed at a distance of 25 cm from the lens on the same side of the object., Q.14 The angle of deviation is decreased., Q.16 (i) +5 D; (ii) 10 cm, , Q.17 25 cm , , Q.20 feq = ∞, , Q.21 (ii) -10 , , Q.23 (a) 5 , , Q.24 2.14, , =, f0 40cm;, =, fe 5cm, Q.25, , Exercise 2, Single Correct Choice Type, Q.1 A, , Q.2 B, , Q.3 A, , Q.4 C, , Q.5 B, , Q.6 A, , Q.7 D, , Q.8 D, , Q.9 A, , Q.10 A, , Q.11 C, , Q.12 D

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P hysi cs | 16.85, , Q.13 B, , Q.14 B, , Q.15 B, , Q.16 C, , Q.19 C, , Q.20 B, , Q.21 C, , Q.22 A, , Q.17 D, , Q.18 D, , Previous Years’ Questions, Q.1 D, , Q.2 A, , Q.3 A → p, r; B → q, s, t; C → p, r, t; D → q, s, , Q.4 A,B,C, , Q.5 6, , Q.6 D, , Q.7 C, , Q.8 B, , Q.9 D, , Q.10 D, , Q.11 B, , Q.12 B, , Q.13 D, , Q.14 D, , Q.15 D, , Q.16 A, , Q.17 D, , (iii) 195°, , (iv) 285°, , (v) 315°, , Q.3 75 cm, , Q.4 2 cm, , Q.5 80 m/s, , (ii)1, , (iii) y = k 2 ( x / 4 ), , JEE Advanced/Boards, Exercise 1, Q.1 (i) 75o, , (ii) 165°, , Q.2 160 cm; 320 cm, Q.8 (i) tan=, θ, , dy, = cot i, dx, , Q.9 5 cm, , (v) It will become parallels to x-axis, , µ12, , + µ32, , Q.10 h=5.95 m, + µ52 =, , 2 + µ22, , Q.13 1000 m/s , , Q.14, , 1, Q.16 (a) sin−1   (b) air, 5, , d, 1, Q.17  , = ,  R max 2, , Q.20, , + µ24, , (iv) 4.0,1;, Q.11 Same, , Q.12 1.625, , Q.15 9 m, 43, 5, , Q.18 45°, , Q.19, , Q.4 D, , Q.5 B, , Q.6 D, , Q.15 B, C, , Q.21 15 cm toward the combination, , 3 , 2, , 4, , Q.7 16 ft, , 2, , (d) 24.56 W / m2, , Q.22 (a) 5000 W / m (b) 0.02 W / m, , (c) 0.214 cm, , Q.23 +60, +4/5 , , Q.24 5.9 cm, 10.9 cm, , Q.25 90 cm from the lens toward right, , Exercise 2, Single Correct Choice Type, Q.1 B, , Q.2 A, , Q.3 C, , Q.7 B, , Q.8 A, , Q.9 A, , Multiple Correct Choice Type, Q.10 B, D, , Q.11 B, C, , Q.12 B, C, , Q.13 A, C, , Q.14 A, B, D, , Q.16 B, D, , Q.17 A, C, , Q.18 A, D, , Q.19 A, C, , Q.20 B, D, , Q.23 D, , Q.24 D, , Q.25 D, , Q.28 D, , Q.29 B, , Q.30 B, , Assertion Reasoning Type, Q.21 C, , Q.22 A, , Comprehension Type, Q.26 B, , Q.27 D, , Q.32 A, , Q.33 A, , Q.31 D

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1 6 . 8 6 | Geometrical Optics, , Match the Column, Q.34 A → p, r; B → q, s; C → p, q, r, s; D → p, q, r, s, , Previous Years’ Questions, Q.1 C, D, , Q.2 B, , Q.3 A → p, q, r, s; B → q; C → p, q, r, s; D → p, q, r, s, , Q.4 A, , Q.5 B, , Q.6 C, , Q.7 B, , Q.8 B, , Q.9 B, , Q.10 3, , Q.11 B, , Q.12 A, C, D, , Q.13 A, , Q.14 A, , Q.15 C, , Q.16 D, , Q.17 A, C, , Q.18 C, , Q.19 B, , Q.20 7, , Q.21 B, , Q.22 2, , Q.23 A, C, , Q.24 A, , Q.25 A, D, , Solutions, JEE Main/Boards, , (ii) Focal length of a concave lens depends on the, refractive index µ of the medium which in-turn, depends upon the wavelength of light. µ decreases with, increasing wavelength. So for red light µ will decrease., , Exercise 1,  A + δm ,  90 , 1, sin ,  sin  , 2, 2, 2, , ,  =, , Sol 1:=, µ, =, =, A,  60  1 / 2, sin, sin  , 2,  2 , o, o, Here A = 60 , δm = 30 ., , Now μ =, , 2, , c, c, 3 × 108, ⇒v= ⇒v=, = 2.12×108 ms-1, v, µ, 2, , 1 1 1, − =. Let object is, v u f, placed at distance x from lens and image is found, , Sol 2: Lens Formula:, , at distance y from lens. For real image v is positive, v +y, v=, + y, u =, − x, m =, ==, −2 (For real image m, u −x, 1 1 1 1, 1, − = −, =, is negative) ⇒ y =, 2x …..(i), (f is, v u y −x 20, positive for convex lens), , 1,  1, 1 , −, As per lens maker’s formula  = (µ − 1) ,   as,  f,  R1 R 2  , µ decrease, f increase., Sol 4: Total internal reflection (TIR) takes place when, light travels from denser medium towards rarer medium, and at the interface the angle of incidence exceeds θc,, the critical angle, and the incident beam is completely, reflected at the boundary (interface). Critical angle, ,  µ, θc =sin−1  Rarer, µ,  Denser, , , , , , 1, 1, 45= sin−1   ⇒ =, µ, µ,  , ⇒ v=, , c, 2, , =, , 3 × 108, 2, , 1, 2, , ⇒ µ=, , c, =, v, , 2, , = 2.12 × 108 ms−1, , Sol 5: Lens formula., , 1 1, 1, 3, 1, 1 1 1, + =, (using (1)) ⇒, =, ⇒ x= =, 30 cm, −, (for concave lens f is ( − ) ve), 2x x 20, 2x 20, v u f, 1 1, 1, − f, −, =, Now for u =, u=-30 cm, v −f −f, 1, 1 1, 2, f, =− − =− ⇒ v =−, Sol 3: (i) Focal length of a concave mirror is independent, v, f f, f, 2, of the medium and wavelength of light. So there will, Image is virtual, diminished and on the same side as, not be any change., object., ⇒

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1 6 . 8 8 | Geometrical Optics, , Sol 10: Refracting astronomical telescope: It consists of, an objective lens of a large focal length (fo) and large, aperture, also an eye lens of small aperture and focal, length., (i) Magnification when final image is formed at D,, ⇒m=, −, , L = |fo| +, , fo , fe ,  1 +  and length of telescope,, fe , D, fe D, , fe + D, , Sol 11: (i) Lens formula, 1 1 1, − =, v u f, u=, +12 cm, f =, +20 cm, 1 1 1 1, 1, 1 1 1, 8, = + =, +, =  + =, v u f 12 20 4  3 5  4 × 15, ⇒v=, +7.5 cm, ⇒, , (ii) f = −16 cm ⇒, , 1 1, 1, 4 −3, 1, =, −, =, =, v 12 16, 48, 48, , ⇒v=, +48 cm, ,  1, 1, 1 , = (µ − 1)  −, Sol 12:, , f,  R1 R 2 , , Sol 14: Angle of minimum deviation δm and angle of,  A + δm , sin , , 2 , , prism A are related as, µ =, A, sin  , 2, , Glass prism of refractive index 1.5 is immersed in a, liquid of refractive index 1.3 so the relative refractive, 1.5, index of the prism decreases. µ=', = 1.15, 1.3, So as per above equation as A is constant for a prism,, as µ decreases, δm also decreases., Sol 15: [Refer question 7 solution], Sol 16: Consider an object O placed at a distance u, from a convex lens as shown in figure. Let its image I, after two refractions from spherical surfaces of radii R1, (positive) and R2 (negative) be formed at a distance v, from the lens. Let v1 be the distance of image formed, by refraction from the refracting surface of radius R1., This image acts as an object for the second surface., Using,, µ2, v, , −, , µ1, , µ2 − µ1, =, u, R, Incident light, , Here R1 = +30 cm ;R 2 = −30 cm; µ = 1.6,  1 , 1, 2, 1, ⇒ = (0.6 )   × 2=, =, ⇒ f + 25 cm, f, 30, 50, 25,  , 1 1 1 1, 1, 1, =, −, u=, 12.5 cm, = + = −, v f u 25 12.5, 25, h2 v, −25, ⇒v=, −25 cm , m = = =, =, 2, h1 u −12.5, , C₂, , O, , Sol 13: Predominance of bluish colour in a clear sky, is due to the phenomena of scattering of light in the, atmosphere around earth. If size of the air particles, are smaller than the wavelength, the scattering is, proportional to 1 / λ 4 . This is the Rayleigh’s law of, scattering. The light of short wavelengths are strongly, scattered by the air molecules and reach the observer., Among the shorter wavelengths, the colour blue is, present in large proportion in sunlight., , 1, , 2, , C₁, , 1, , I, , +ve, u, , ⇒ h2 =2h1 =2 × 5 cm =10 cm, , Image is virtual and erect, on the same side as the, object., , R₁, , R₂, , twice, we have, and, , µ1, v, , −, , v, , µ2, v1, , −, , µ1, , µ2 − µ1, , =, ....(i), u, R, , µ2, , µ1 − µ2, ....(ii), =, −R 2, v1, , ... (i), ... (ii), , Adding Eqs. (i) and (ii) and then simplifying, we get, , 1 1, − =, v u, ,  µ2,  1, 1 , − 1 , −, ,  .....(iii), ,  µ1,  R1 R 2 , , ... (iii), , This expression relates the image distance v of the, image formed by a thin lens to the object distance u, and to the thin lens properties (index of refraction and, radii of curvature). It is valid only for paraxial rays and

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P hysi cs | 16.89, , only when the lens thickness is much less then R1 and, R2. The focal length f of a thin lens is the image distance, that corresponds to an object at infinity. So, putting, u=, ∞ and v =, f in the above equation, we have, ,  1, 1  µ2, 1 , =, − 1 , −, ,  ....(iv), , f  µ1,  R1 R 2 , , Sol 19: (a) (i), Q, , P, , This is called the lens maker’s formula because it can, be used to determine the values of R1 and R2 that are, needed for a given refractive index and a desired focal, length f., Combining Eqs. (iii) and (v), we get, 1 1 1, − = ....(vi) which is known as the lens formula., v u f, , (i) P = P1 + P2 = 10D − 5D = 5D, , 1, 1, v, (ii)f = = =0.2m =20 cm; m =, + 2=, P 5D, u, ⇒v=, 2u, For virtual image m is positive, 1 1 1, 1 1 1, = + ⇒, − =, v f u, 2u u 20, 1, 1, ⇒−, =, ⇒ u = −10 cm, 2u 20, , E, , , , ’, , O, , O’, , If the refractive index of the material of the lens is µ, and it is placed in air, µ2 =µ and µ1 =1 so that Eq. (iv), becomes, ,  1, 1, 1 , = (µ − 1) , −,  ....(v), f,  R1 R 2 , , P’, , , , P’’, , Q’’, , (ii) Reflector telescope advantages:, 1. Reflector telescopes do not suffer from chromatic, aberration because all wavelengths will reflect off the, mirror in the same way., 2. Support for the objective mirror is all along the back, side so they can be made very BIG., 3. Reflector telescopes are cheaper to make than, refractors of the same size., 4. Because light is reflecting off the objective, rather, than passing through it, only one side of the reflector, telescope’s objective needs to be perfect., (b) f1 = +10 cm, f2 = -10 cm, u = -30 cm., 1, 1 1, 1, 1, 2, = +, = −, = ⇒ v1 =15 cm, v1 f1 u1 10 30 30, , So for the concave lens, u2 =, +(15 − 5) cm =, +10 cm, f0, , B, , fe, , Sol 17:, , 1, 1, = 0.25m, P = P1 + P2 = 6 − 2 = 4D ⇒ f = =, P 4D, ⇒f=, 25cm, Sol 18: The speed of light in vacuum is a universal, constant denoted by c. When a light wave travels in a, transparent material, the speed is decreased by a factor, µ , called the refractive index of the material., , µ=, , speed of lightin vacuum, speed of light in the material, , For graph refer figure of question 6., , A, , Fo, A’ F’e, , O, , Fe, E, , B’, , , , 1, 1 1, 1, 1, =+, = +, =, 0 ⇒ v2 =, ∞, v 2 f2 u2 −10 10, , Final image will be formed at infinity., Sol 20:, , 1 1 1 1 1, = + = − =0 ⇒ F =∞, F f1 f2 f f, , Sol 21: (i) Astronomical Telescope for normal, adjustment, It consists of two converging lenses placed coaxially.

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1 6 . 9 0 | Geometrical Optics, , The one facing the distant object is called the objective, and has a large aperture and a large focal length. The, other is called the eyepiece, as the eye is placed close, to it. It has a smaller aperture and a smaller focal length., The lenses are fixed in tubes. The eyepiece tube can, slide within the objective tube so that the separation, between the objective and the eyepiece may be, changed., Eyeplece, vo, vo, , ue, , fo, , h, , F, h’, , ’, , Objective, , Image, , When the telescope is directed towards a distant object, PQ, the objective forms a real image of the object in its, focal plane. If the point P is on the principal axis, the, image point P’ is at the second focus of the objective., The rays coming from Q are focused at Q’. The eyepiece, forms a magnified virtual image P”Q” of P’Q’. This, image is finally seen by the eye. In normal adjustment,, the position is so adjusted that the final image is, formed at infinity. In such a case, the first image P’Q’ is, formed in the first focal plane of the eyepiece. The eye, is least strained to focus this final image. The image, can be brought closer by pushing the eyepiece closer, to the first image. Maximum angular magnification is, produced when the final image is formed at the near, point., f, p, 10D, − 0 =, − e =, −, =, −10, (ii) m =, fe, pο, 1D, , Here the objective has large focal length and smaller, Power., (b) (i) Figure shows a simplified version of a compound, microscope and the ray diagram for image formation., It consists of two converging lenses arranged coaxially., The one facing the object is called the objective and, the one close to the eye is called the eyepiece or ocular., The object is placed at a distance u0 from the objective, which is slightly greater than its focal length f0 . A real, and inverted image is formed at a distance v 0 on the, other side of the objective. This image works as the, object for the eyepiece. For normal adjustment, the, position of the eyepiece is so adjusted that the image, formed by the objective falls in the focal plane of the, , eyepiece. The final image is then formed at infinity., It is erect with respect to the first image and hence,, inverted with respect to the object. The eye is least, strained in this adjustment as it has to focus the parallel, rays coming to it. The position of the eyepiece can also, be adjusted in such a way that the final virtual image is, formed at the near point. The angular magnification is, increased in this case. The ray diagram in figure refers, to this case., (ii) Magnifying power of a Compound microscope is, , v D , m=, − 0   → normal adjustment and, u0  fe , v , D, m=, − 0  1 +  → final image at D., u0 , fe , Now for large magnification, m is to be large, so fe, should be small and u0 should be small. Now object, is placed at a distance u0 from the objective which is, slightly greater than its focal length fo. So for u0 to be, small, fo should also be small., Sol 22: Refer question 19.(a).(ii), Sol 23: When two plane mirrors are placed at an angle, θ to each other, the object is kept between them, then, 360ο, the numbers of images observed is n =, . If n is, θ, even then number of image is (n-1)., So here, , 360ο 360ο, =, = 6⇒ n = 6 − 1 = 5, 60, θ, , (b) At each reflection some of the light energy is lost, due to absorption at the mirror surface. So the intensity, of the reflected ray goes on decreasing at multiple, reflections due to parallel mirrors., , v D , Sol 24: For compound microscope m = − ο  , uο  fe , (normal adjustment),  v, −20 =  − ο,  u,  ο, ,   25 cm , v ο −60 −12, =, =, = −2.4,  ,  ⇒−, uο, 25, 5,   3cm , , For final image at least distance:, , v , v , D, 25 , m=, − ο  1 +  ⇒ − 20 = − ο  1 +, , , , uο , fe , uο , 3 , ⇒−, , vο, , uο, , =, , v, −20, −5 × 3, =, ⇒ − ο = −2.14, 28 / 3, 7, uο

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P hysi cs | 16.91, , −fο, , m, Sol 25: Astronomical telescope =, adjustment, fo, , ⇒ −8 = −, , fe, , fe, , → normal, , left with speed 3v. So its second image will more away, towards right with speed 3v w.r.t mirror 2. Hence speed, w.r.t O will be 3v+v=4v, Hence for nth image v image = 2nv, , ⇒ fo = 8fe, , L = fο + fe = 45 cm ⇒ 8fe + fe = 45 cm, , ⇒ 10h + y= 3y ⇒ y = 5h, , Exercise 2, , For general case, , L₁, , Single Correct Choice Type, , M₂, , 3h, , Sol 1: (A), , tan θ =, , o, , N, , , , , ’, ’, , 1, , ⇒ 2., , N’, , o, , o, , 70, , 2, , , , ο, , α = 180o − 70o − (90o − θ ') = 180o − 70o − 70o, , , y, , S, , 3h, h, =, ⇒ 3y = x + y ⇒ 2y = x, x +y y, , dy dx, dy 1, =, ⇒, = .v, dt dt, dt 2, , Sol 4: (C) A is head and E is feet of man. C is the eye., The mirror can be placed anywhere between the centre, line BF (of AC) and DG (of CE) to get full image from, head to feet., So here CE=1.4 m. So DE should be 0.7 m. But mirror is, 0.8 m from ground so feet will not be visible. The upper, edge of mirror is at height (0.8+0.75) m equal to 1.55 m, which is more than BE., , =, α 40o ⇒=, θ 90o −=, α 50o, , A, , Sol 2: (B) With respect to mirror1 the object is going, array from mirror. So first image will also more away, w.r.t mirror 1 with same speed v. So with respect to, object O the image speed is, , 1, , v, , 10h = x, , 70, , 90 -’, , M₁ h, , L₂, , =, δ1 180o − 2θ; =, δ2 180o − 2=, θ ' 70o + 90o − θ ', , ⇒ θ ' =20, , h, 3h, lim, =, y 10 h + y x →∞, , Sol 3: (A) tan θ=, , =, ⇒ fe 5cm=, and fe 40 cm, , 2, , x, x, , B, C, , D, y, E, Man, , AC, 0.1, + CE =, + 1.4, 2, 2, = 0.05 + 1.4 = 1.45m, , BE = BC + CE =, , Now this image becomes object for mirror 2. With, respect to mirror 2 the image is going away towards, , (x + y), , y, , O, , v image1 = v image1,mirror1 + v mirror 1,obj = v + v = 2v, , F, , So head will be visible., , G

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1 6 . 9 2 | Geometrical Optics, , Sol 5: (B) Rays from S going right from the normal will, not reach the bottom horizontal mirror as they will hit, the inclined mirror and get deviated. So C will see only, the image formed by inclined mirror., , When the mirror is present, , Sol 6: (A) Object O moves towards M1 so image 1 due, , Ratio =, , For M2 we have formula for speed of image as, , P, , to M1 will move towards left i.e. towards M2 ., , Sol 7: (D) Intensity incident, =, , Pθ2, 4 × (area on which lightis incident), , When the mirror is not present, light is reaching the, screen up to height h. Maximum area on which light is, incident = πh2, Intensity =, tan θ =, , Pθ2, 4 πh2, , 4 πh1, , 2, , and tan θ =, , Intensity when mirror present, Intensity when mirror object, , P, +, 4 π × (20), 4 π × (60)2 10, =, P, 1, 4 π × (20)2, , Sol 8: (D) As a result of water the apparent height of, source will be beyond C, at C’. OC’ = Rμ. So its image I’, will not be formed at C but it will be formed between, C and focus F of the mirror. But again in the return, path of rays they will be again refracted at water to, air boundary and final image I will be further shifted, downwards towards O., , C, , and, , I’, I R, F, , 2, , h, Ph, P, ⇒, =, =, 2, 2, 60, 4 πh × (60), 4 π × (60)2, , , , , , O, , , Sol 9: (A) The slab will cause a lateral shift in the, incident rays as well as in the reflected rays from the, circular mirror MM’. Now the angle of emergence θ1, will be equal to the angle of incidence in case of a slab., M, , h, s, , h, P, ⇒=, 20, 4 π × (20)2, , 2, ,  v 2  du, =, dv, = −   . So negative sign means final image,  u2  dt, dt,  , 2 due to M2 will move opposite to image 1 i.e. towards, right., , Pθ12, , Intensity =, , , , , C, , 60, , 1 1, , M’, B, , h, A, , D, P, , h, , 20, , F, , 1, , Q, , , , O, , The rays reaching the edge M of the circular mirror, after passing through the glass slab will be leaving the, source O at a greater angle (θ1 ) with the normal as, compared to the angle (θ) when there is no slab. But, due to symmetry of incident and reflected rays, the, reflected rays from the edge M, after passing through, the slab will reach the some point Q on the ground, where they would have reached when there was no, slab.

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P hysi cs | 16.93, , Here we have, OQ=QP, both without and with slab,, between source O and mirror MM’., , Sol 10: (A), , B, , r, , h, , E, , c, , , 90 -, o, , O, n=, , , 3, 2, , A, , n=, C, , 6 /5, 90 − θ ≥ θc = sin−1 , ;, 3/2, 4, 90 − θ ≥ sin−1   ;, 5, ο, , 90 − θ ≥ 53 ⇒ θ ≤ 37, , Ω = 2π(1 − cos θc ), , 6, 5, D, , =, I, , I, .2π (1 − cos θc ), 4π, , I' (1 − cos 48.59 ), I, =, ⇒, =, ⇒, 16.9 %, I, 2, I, , ο, , Sol 13: (B), , Sol 11: (C), 1, h 4m ; θc = sin−1  , =, µ,  1 , −1  3 , ο, = sin−1=, =, ,  sin,   37, 5 / 3, 5, tan θc =, , If total intensity is I then, intensity per unit solid angle is, I, .So intensity through the circular area is,, 4 π., , r, 3h 3 × 4m, ⇒ r = htan 37ο =, =, h, 4, 4, r, r, , =, v1, , =, and m2, , c, , c, , f u1, f u2, =, and v 2, u1 − f, u2 − f, , Now m1=, , h, , ⇒ diameter =, 2r =, 6m , , 3, −1 1, ο, sin−1=, =, θc sin=, Sol 12: (D),   48.59, 4, µ,  , Solid angle subtended at source of light O by the, circular area of radius r is, , v1, −f, v, =, =, u −u1 u1 − f, −f, =, .also | m1 | | m2 |, u2 − f, , Now m1 is negative (real image) and m2 is positive, (virtual image). So we have,, −f, f, =, ⇒ u2 − f =−u1 + f, u1 − f u2 − f, ⇒, , O, , 1 1, 1, 1 1 u1 − f, = +, = −, =, v1 f ( − u1 ) f u1, f u1, , u1 + u2 = 2f ⇒ f =, , u1 + u2, 2, , , , Sol 14: (B) The image formed by first lens will lie at its, second lens focus. This image will act as an object for, the second lens. For the rays to become parallel after, passing through the second lens, the object for second, lens should lie on its first focus. Thus the distance, between the two lenses will be equal to sum of their, focal lengths., D = f1 + f2 = 20 cm + 10 cm = 30 cm

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1 6 . 9 4 | Geometrical Optics, , Sol 15: (B) Image formed by lens be at distance v1 from, lens., , 1, 1 1, 1, 1, 3−2 1, = +, = +, =, =, v1 f1 u1 10 −15, 30, 30, , Speed of image with respect to object is, = v I,L + vL,O, v I,O, ,  dv , =, + ( − v), v I,O  , (towards right),  dt (towards left), , v1 = 30 cm from lens., , Sol 18: (D) At first refracting surface we have,, sini1 = µ sin r1 . So as i1 decreases, r1 also decreases., , O, 10 cm, , For convex lens, u2 = +(30 - 10) cm = +20 cm, v2= +20 cm, because rays retrace their path after, reflection., 1, 1, 1, 1, 1, 1, 1, +, =⇒ = +, = ⇒ f2 =, + 10 cm, v 2 u2 f2, f2 20 20 10, , Now for prism r1 + r2 =A (constant). So as r1 decreases,, r2 increases. At the second refracting surface we have,, µ sin r2 =, sin i2 . So as r2 increases, i2 also increases. So, out of all choices D is most appropriate as amount of, increase in i2 should be less than amount of decrease, in i1., Sol 19: (C) For prism with refracting angle A, we have, ,  A + δm , sin , , o,  2=,  ⇒ 3 sin (90 + δm ) / 2, =, µ, Sol 16: (C) For first lens, convergent ray becomes, A, 2, sin 45o, sin, parallel to principal axis after refraction., 2,  90o + δ  90o + δ, So f1 = +5 cm., 3, m , m, =, ⇒, ⇒=, sin , 60 o, , , 2, 2, 2, For second lens, ray parallel to principal axis becomes, , , convergent and parallel to incident ray., ⇒=, δm 120o −=, 90o 30o , , , h, , , , h, , , , X, 5 cm, , So focal will length of second lens will be x as shown, in figure., tan θ =, , h h, =, ⇒ x = 5 cm ⇒ f2 = x = 5 cm, 5 x, , Sol 17: (D) Let as work in the frame of reference, 1 1 1, attached to the lens. Lens formula: − =, v u f, Differentiating w.r.t. time,, dv, du, − v −2, − ( −u−2 ), =0 ; (f is constant), dt, dt, , ⇒, , dv  v 2  du, =,  , dt  u2  dt, , Initially when u=f, v → ∞ so speed image is very large, and finally when u → ∞ , v → f and the speed of image, is very low (nearly zero). With respect to lens, as object, moves left, the image also moves left., , Sol 20: (B) At min deviation, =, im, , A + δm A + 38ο, =, 2, 2, , Now δ= (i1 + i2 ) − A; Here i1= 42o, and=, i2 62ο=, , δ 44o, ⇒ 44 ο= (42ο + 62ο ) − A ⇒ A= 104 ο − 44 ο= 60o, =, ⇒ im, , 60o + 38o, = 49ο , 2, , Sol 21: (C), , I, , x O, , , M, , P, , δ = (µ − 1) A; µ = 1.5, o, =, δ 0.5 × 5=, 2.5o ⇒ =, δ, , 2.5, 5, π, =, π, =, π, 180, 360, 72, , OP ≈ OM ≈ IM = 10 cm;, , ∆x = δ × (OM) = 10 δ cm

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P hysi cs | 16.95, , ∆x=, , Sol 4: (A, B, C) Using Snell’s law, , 5π, π, × 10 cm=, cm, 72, 36, , B, O, , Sol 22: (A) At second refracting surface, , o, , 60, , o, , 135, , P, , A, , µ sin r2 =, 1sin 90, 1, 1, ⇒ sin r2 ==, µ, 2, ⇒ r2 =, 45ο, , C, , ο, , = 60 − 45 = 15, , 45, , o, , 60, , , , i, , r1, , r2, , 90o, o, , ⇒ r1 = A − r2, ο, , o, , A, , = 2, , ο, , 30, , 75o, , ⇒ At first refracting surface , sin i1 =, µ sin r1, i1, ⇒ sin, =, , 2 .sin =, 15ο, ,  3 −1, sin−1 , ⇒ i1 =, .,  2 , , , , 2., , 3 −1, =, 2 2, , 3 −1, 2, , Previous Years’ Questions, Sol 1:, , 1, 1, 1, – = = constant, f, v, u, , Sol 2: (A) An experiment is performed to find the, refractive index of glass using a travelling microscope., In this experiment, distances are measured by a vernier, scale provided on the microscope., Sol 3: (A) → since µ1 < µ2 , the ray of light will bend, towards normal after first refraction., , 1, , sin–1, , 3, , < sin–1, , 1, 2, , Net deviations is 90º, Sol 5:, 1 1 1, − =, v u f, , or, , u, u, − 1 = or, v, f, , ∴, , v  f , m ==, , , u u+ f , , u u+ f , =, , v  f , ,  20 , m25  −25 + 20 , =, = 6, m30  20 , , ,  −50 + 20 , , (B) → µ1 > µ2 , the ray of light will bend away from the, normal after first refraction., , ∴ Answer is 6., , (C) → µ2 =µ3 means in second refraction there will be, no change in the path of ray of light., , Sol 6: (D) Case I: u = – 240 cm, v = 12, by lens formula, , (D) → Since µ2 > µ3 , ray of light will bend away from the, normal after second refraction., , 1, 7, =, f, 80, , Therefore the correct options are as under., , Case II: v = 12 –, , (A) → p, r, (B) → q, s, t, (C) → p, r, t, (D) → q, s, , 1, 35, =, 3, 3, , (normal shift = 1 –, f=, , 7, 80, , u = 5.6, , 2, 1, = ), 3, 3

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1 6 . 9 6 | Geometrical Optics, , Sol 7: (C) L.C =, , 1, 60, , 1, =, f, , 9, Total Reading = 585 +, = 58.65, 60, , Sol 8: (B) As intensity is maximum at axis., \ µ will be maximum and speed will be minimum on, the axis of the beam., ∴Beam will converge., Sol 9: (D) For a parallel cylindrical beam, wave front will, be planar., Sol 10: (D) Case I: u = -240 cm, v = 12, by Lens formula, , 1, 7, =, f 80, Case II: v = 12 −, , 1 35, =, 3 3, , , 2 1,  Normal shift = 1- = , 3 3, , f=, , 7, 80, , , , ( µ − 1)  R1, , , 1, , 1 , , R 2 , ,  1 , 1 3, =  − 1  , f 2,   15 , f = 30 cm, Sol 13: (D), , fm, f, , =, , (µ − 1),  µ, , − 1 , , µ,  m, , , 3, ,  − 1, f1, 2,  =, , =, ⇒, 4, f 3/2, , − 1, , 4/3, , ⇒ f1 =, 4f, , 3, , − 1, , f2, 2,  = −5, = , f 3/2, , − 1, , 5, /, 3, , , ⇒ f2 < 0, , u = 5.6, Sol 11: (B) Self-explanatory, Sol 12: (B), R 2 = d2 + (R − t ), , 2, , 2, , , t, R 2 − d2= R 2 1 − , R, , , , R, , Sol 14: (D) As frequency of visible light increases, refractive index increases. With the increase of, refractive index critical angle decreases. So that light, having frequency greater than green will get total, internal reflection and the light having frequency less, than green will pass to air., Sol 15: (D) At face AB,, sin θ = µ sin r , , At face ACr ' < θc, , d, A − r < sin−1, , R-t, , t, , 1, µ, , ∴ r > A − sin−1, , 2t, 1−, =, 1−, 2, R, R, , (3) =, 2 × ( 0.3 ), 2, , 90, = 15 cm, 6, , , B, , 1, µ, , , 1, ∴ sin r > sin  A − sin−1 , µ, , , d2, , =, R, , −, , , sin θ, 1, > sin  A − sin−1 , µ, µ, , , , , , 1 , θ > sin−1 µ sin  A − sin−1  , µ , , , , A, r, , r’, C

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P hysi cs | 16.97, , Sol 2:, , Sol 16: (A) θ =1.22 λ, D, , (, , ), , θ 30 µm, Minimum separation = 25 × 10−2 =, Sol 17: (D) δ = i + e − A ⇒ A = 74°, , 120 cm, ,  A + δmin , sin , , 2, ,  5 sin  37° + δmin , =, µ, =, , , 3, 2 , A, , sin  , 2, µmax can be, , A, , 40 cm, , 40 cm, B, , 5, 5, , so µ will be less than, 3, 3, , , , , F, C, 40 cm, , 40 cm, , , , Since δmin will be less than 40° , so, , µ<, , P, S, , 5, 5, sin 57° < sin 60° ⇒ µ < 1.446, 3, 3, , 20 cm, , So the nearest possible value of µ should be 1.5, , 40 cm, , By similar triangles, ∆ BCD  ∆ BFE; so EF =, 3 × CD, Because BF= 3 ×BC, ⇒, =, EF 120 cm ⇒, =, EP 160 cm, , JEE Advanced/Boards, , ∴ Minimum height of eye is 160 cm., , Exercise 1, , And similarly maximum height will be,  3 −1, 2=, , ,  2 2 , , , , Sol, 1: OB OP, sin15, =, =, ο, , E'P = 80 + 3 × 80 = 320cm, , 3 −1, 2, , Sol 3: By property of similar triangles,, , Number of image, y, , M, , P, , B, , o, , (1, 1), , A, (1, 0), (1, -1), , 360, 360, − 1=, − 1= 5, θ, θ, , 3m, S, , B, , o, , P, (0, 0), , x, , x, , 120, 60, , 20 cm, ,  , , O, , n, =, , G, , D, , 20, , 20, , x, , c, , 3-x, , D, , I1, , ∆ MAB  ∆MCD;, , 300 cm, x, =, 20 cm, y, , y=, + 20 100 cm;=, ⇒ y 80 cm, 20 × 300 cm, =, ⇒x, = 75cm, 80

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1 6 . 9 8 | Geometrical Optics, , Sol 4:, M1, axis of M₁, , 10 cm, , dv, =, dt, , I₂, 1cm, , A, , O, , 1cm, , Relation between speed of image and speed of object, for lens is, , 1cm, ,  v 2  du, dv, ⇒, =,  2 .,  u  dt, dt,  , ,  ( −30)2 , ,  . 20m s−1 (downward),  ( −15)2 , , , , dv, (4) × 20 ms−1 =, 80m s−1 (downward), =, dt, , 1cm, , I₁, axis of M₂, M2, , 1 1 1, + = ;f =, −20 cm ,u =, −10 cm, v u f, 1 1, 1, 1, 1, 1, 2 −1 1, −, =, −, ⇒ = −, =, = ⇒v=, 20, v 10, 20, v 10 20, 20, 20, v, 20, M=, =, −, =, 2, u, ( −10), , Image will be erect with respect to the axis of each, mirror. Distance between images is 2 cm., , Sol 6: Refer theory, Sol 7: Apply Snell’s law:, , 4, sin(90 − ϕ), 3, d, 4, d×3, 1×, =, ×, 2, 3, 30 + d, 1024 + d2, , 1=, × sin θ, , 1024 + d2= 4 36 + d2, , Sol 5: OS=15 m; OC =20 m, Height of ball after 4 s is,, , 2, , 6 ft, , , C, Speed s = 20m/s, , 2, , 6 +d, , S, , , Vertical shaft, 1 feet = 12 inch, 1' = 12", , , d, , S, , d, 90°-, 10, 2, , 1024+d, 9, , 1, H = 15 + 20 × 4 − × 10 × 16 = 15 + 80 − 80 = 15m, 2, Speed of ball after 4 s is, sball = 20 − 10 × 4, ⇒ sball =, 20 − 40 =, −20 ms−1, So ball is moving downward with speed of 20 ms-1 and, is at height 15 m above the mirror., So u=-15 m, and f =, Mirror formula, , 20, m = −10 m, −2, , 1 1 1, + = gives, v u f, , 1, 1, 1, 1, 1, 2 − 3 −1, =, −, = −, =, =, v −10 ( −15) 15 −10, 30, 30, ⇒v=, −30 cm, , 4”, o, , 1', 3, , On squaring both sides,, 1024+9d2=16(36+d2) ; 1024+9d2=576+16d2, 7d2=448; d2=, , 448, = 64, 7, , d= 8 feet, width=2d=16 feet, Sol 8: At any point by Snell’s law, 1sin 90 =n(y). sin (90o − θ);, 1=, (kg3/2 + 1)1/2 cos θ, , 2', 3

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P hysi cs | 16.99, , Here angle of incidence at B(x, y) is i = 90° – θ, , dy, [From (1), (i) put, put yy =, 1], = tan θ = 1 ⇒ θ = 45ο [From, = 1], dx, , y, , n =+, (1 1)1/2 =2 ⇒, , r, , i, , ⇒ sin r = 1 ⇒ r = 90ο, , B(x, y), , ⇒ Ray of right becomes parallel to x axis., x, , (0, 0), , cos θ =, , 1, (ky, , 3/2, , Sol 9: At first refracting surface the rays will pass undeviated. At the second refracting surface the rays are, refracting from denser to rarer medium and hence, suffer Total Internal Refraction if i > θc, , + 1)1/2, , dy 1, = =, dx 2, , ⇒ tan θ=, , sin 45o, 1, =, sin r, 2, , ky 3/2 + 1 − 1, 1, , =, , ky 3/2, , B, c, , (i) Now as, dy, =, i 90o − θ ⇒=, θ 90o − i So = tan=, θ cot i, dx, , R, , (ii) Initial angle at air is 90° and n=1. At point B(x, y), angle of incidence is i., , c, , So we have by Snell’s law, , A, , 1.sin 90ο = n sin i, ⇒ n sin i =, 1, , (iii) Now,, y, y, , 0, , x, x, , 4, 1/ 4, yy ++1/, kk dx, ⇒, dx ⇒ +1 / 4 ==, +1 / 4, , ∫∫, 0, 0, , 90-c, D, , P, , C, x, ,  3, 1, =, θc sin−1  =, ⇒ θc sin−1=, ,  60ο,  2 , µ, , , , dy, (i)], = k y 3/ 4 [From, [From(1)], dx, , dy, dy, ⇒, ⇒ ∫∫ 3/ 4 ==, 0y, y 3/ 4, , h, , From right ∆ BCD, ∠BCD, = 90o − θc, , kk xx, , o, From ∆ ABD,=, h R sin θ=, R sin 60=, R, c, , 4, 4, , 3, 2, , x, kk xx, yy kk 22  x  …(ii), .......(2), ⇒, =, ⇒, =,  4  .......(2), 4, 3, 3R, 4, 4, o, o,  , =, − θc ) h cot30, =, DC h cot (90=, R=, . 3, 2, 2, (iv) For the point P, we have y=1.0 m k=1, R, = R , AD, = R cos=, θc, AP, 4, 4, 2, x, 2x, ⇒y=, k   gives 1 =   ⇒ x = 4.0 m, R 3R, 4, 4, =2R;, AC =AD + DC = +, 2 2, (v) At P, we have, x = pc = AC − AP = 2R − R = R = 5cm, +1/ 4, 4, ⇒, ⇒ yy +1/=, =, , ⇒x=, 5m, , y, (0, 1), , P, , r, , (2, 1), 45o, , x

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1 6 . 1 0 0 | Geometrical Optics, , Sol 10:, , For reflection of reflected ray at first face of prism, , o, , 37, , A, r, , r, r, , h, , i, =1 15', o, , (90-2r), 2r, 3.0 m, , o, , 6 30', , ο, , sin i, sin 37, 4, =µ⇒, = µ = ⇒ sin r =, sin r, sin r, 3, ⇒, , 3, , =, , 2, , 3/5, 4/3, , h +9, 1, 3, 400, =, ⇒ 400 = 9 (h2 + 9) ⇒ h2 =, −9, 2, 30, 9, h +9, , h, ⇒, =, , 6.5, µ sin 2r =sin 6.5 = π, 180 1, , 319, = 5.95 m, 9, , ⇒=, 2rµ, , sin i, = µ All the ray in the bean are deviated, Sol 11:, sin r, by same angle so width of beam will not change after, it goes over to air., , i, , , , , , From (i), (ii) & (iii) we get, 1.25 π, = (µ − 1) A = µ A − A .....(4), 1.25, 180π = (µ − 1) A = µ A − A .....(4), 1806.5 π, and 6.5 π = µ A.....(5), , and 360 = µ A.....(5), 360, , =, µ, , ο, , Sol 12: δ = (µ − 1) A = 1 15' = 1.25 = (µ − 1) A , , … (i), , A. Here r1 = 0,, Now for prism r1 + r2 =, r2 = r, ⇒ r =A , , … (v), , and put value of A in (v) to get, , r, , ο, , … (iv), , (6.5 − 2.5)π, π, = A⇒A =, rad = 2 ο, 360, 90, , Glass, Air, , r, , … (iii), , Subtract (iv) from (v) to get, , ii, , L, , 6.5π, 6.5π, , ⇒=, rµ, 180, 360, , ... (ii), , 6.5 π 90, ×, ⇒, =, µ 1.625, 360, π, , Sol 13: For a given incident ray, if the mirror is rotated, through an angle θ , then the reflected ray turns, through an angle of 2 θ . So if angular speed of mirror, is ω then the angular speed by which the reflected ray, is rotated is 2 ω., ωrefl = 2 × ω = 2 ×, 36 rad s−1, ωrefl =, , 9 r e v 18, rad, =, × 2π, sec, π sec, π

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1 6 . 1 0 2 | Geometrical Optics, , Sol 17: The outer most ray of the beam , ray 1, will, be tangential to the circular surface of rod at point P, and hence angle of incidence is 90°, hence greater than, critical angle, and hence will travel tangentially at all, points of the circular portion from P to P’., i, , Sol 19: Angle of deviation is gives as δ= (i1 + i2 ) − A, Hence i2 – i1 = 23°, A =60°, δ =23°, ⇒, 23°°==, ⇒, ⇒ 23, 23°=, ⇒, ii1 +, i, ⇒, ⇒ i11 +, + ii222, ⇒, ⇒ iii22 −, − ii11, ⇒, 2 − i1, , ii1 +, i − 60, i11 +, + ii222 −, − 60, 60, == 83, °, ...(1), , 83, °, ...(1), = 83° ...(1), == 23, 23°°°  ...(2), ...(2), = 23, ...(2), , … (i), … (ii), , From (i) and (ii) we get i1 = 30°, i2=53°, Snell’s law at first refracting surface., , R, P, , Q, , Q’, , P’, o, , 60, , d, A, , i1, , B, , ray 1 ray 2, , 1, , 2, , o, , From the geometry of figure we see that,, CQ = R; CA= R +d, Here angle i will be the least of all angles of incidence, of ray 2 during its path inside the critical rod. So, if i is, greater than critical angle then ray 2 will surfer TIR at all, point in circular rod., , ⇒, , o, , sini1 =, µ sinr1 ⇒, , 60, , 1, 1, =, µ sinr1 ⇒ sinr1 =, 2, 2µ, , Snell’s law at second refracting surface., , R=(R+d) cos (90 - i) ⇒ R = (R + d) sin i, , 1, R, 1, ⇒, > ⇒, µ, R+d µ, , 1, 1+, , d, R, , >, , µ sinr2 = sini2 ⇒ sinr2 =, , 1 4, 4, . =, µ 5 5µ, , Now r1+r2=60°, , 1, d, ⇒1+ < µ, µ, R, , A, i, R+d, , d, d, < (µ − 1) ⇒  , = (µ − 1) = 1.5 − 1, R,  R max, , d, 1, ⇒ , =,  R max 2, , 90-i, Q, , Sol 18: In a prime r1+r2=A;, 30o , Here r =r, r =0o, A=30o ⇒ r =, 1, , i2, , r2, , 60, , The inner most ray of the beam, ray 2, will be incident, on the inner circular surface at angle i., , sini >, , r1, , 2, , At first refracting surface, , 1, sin i, 1, = µ = 2 ⇒ sin i = (sin30o ) 2 = 2   =, sin r, 2, 2,  , , ⇒ i =45, , …..(i), , R, , 3, ⇒ sinr1 cosr2 + cosr1 sinr2 =, 2, 1, 16, 1 4, 3, ⇒, + 1−, =, . 1−, ., 2, 2, 2µ, 2, 25µ, 4µ 5µ, ⇒ sin(r1 + r2 ) =, , =, , 43, 5, , C, , 3, ⇒ μ, 2

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1 6 . 1 0 4 | Geometrical Optics, ,  µ1 − 1   1, 1 , −, ,  , , −, 1, 40, 40 , , , 1, 1, ⇒ ( µ w =4 / 3) ⇒ PL = ⇒ PM =−, 60, fM, 1, 1, =, where, fL, fL, , =, PL, , Where, , 1, R, = − (for concave mirror), fM, 2, , 1, R, = − (for concave mirror), fM, 2, PM=, , 2, 2, =, R 40, , −, , 1, 2, 5, 1, +, =, =, 60 40 60 12, , 1, ⇒ fe =, −12 cm, fe, , Again apply, , ⇒, , Iimage =, , 0.02 × π (7.5)2 × 10−4 W, 2, ,  0.214 , −4 2, 4π ,  × 10 m,  2 , , s, , Iimage = 24.56 Wm−2, , Sol 23: Focal length of Plano - concave lens, 1 1, 1, = (µ1 − 1)  − , f1, ∞ R, , ⇒ Pe = 2PL + PM, 2×, , (d) Light intensity at image is the intensity focused by, the lens., , 1 1 1, + =, v u fe, , 1, 1, 1, + = ⇒ u =−15 cm, −60 u −12, ,  1 , 1 3, 1, =, −, ⇒ f1 =, −60 cm,  − 1 −  =, f1  2, 30, 60, , , , Focal length of plano-convex lens, 1, 1, 1 , = (µ2 − 1)  −, , f2,  ∞ −R ,  1 , 1 5, 1, =  − 1  =, ⇒ f2 = 120 cm, f1  4, 30, 120,  , , (i) Plane surface of Plano-convex lens is silvered. So the, equivalent focal length of the system,, , 1, 1 1, 1, 1 1, = + +, + +, −F f1 f2 −fm f2 f1, , When air is filled between the gap, object distance=30, when water is filled between the gas, object distance=15, cm, Then, object is displaced by 15 cm towards the, combination., , Power =, , Sol 22: (a) At surface light intensity, , ⇒ +F = +60 cm, , 4.5 πW, 1.125 × 10 4 4 −2, 4.5 πW ⇒ I, 1.125 × 10, I, =, Wm, −2, 4, 4 2, =, I 4 π.(1.5)2 4.5, =, ⇒ I 1.125, π−W, 2.25 × 10 Wm−2, ×, 10, m, −, 2, 4, 2, 2.25, =, I, =, ⇒I, Wm, 4 π.(1.5), ×, 10, m, −22 × 10 −4 m2, 2.25, 4 πWm, .(1.5), = 5000, = 5000 Wm−−22, = 4.5, 5000, Wm, πW, 4.5 πW 0.02Wm−2 −2, =, IP, =, 2, =, =, 0.02Wm−2, (b) IP4 π.(7.5), 4.52 m, πW, =, IP 4=, π.(7.5)22 m22 0.02Wm, 4 π.(7.5), m, u =, −7.5m, =, −750cm, u =, −7.5m =, −750cm, −7.5m =, −750cm, (c), f =u +=, 30cm, f = +30cm, f = +30cm, , fm = focal length of planemirror = ∞, ⇒, , 1, 2 2, 2, 2, −1, =+ =, −, +, =, −F f1 f2, 60 120 60, , Focal length = 60 cm, The equivalent system behaves as a convex mirror., 1 1 1, (ii) Mirror formula, + =. Hence u = -15 cm,, v u f, f = +60 cm, , 1, 1, 1, 1, 4 1, =, −, =, +, =, v 60 −15 60 60 2, ⇒v=, +12cm, v −12, 4, − = ⇒m=, Magnification, m =, u −15, 5, ⇒, , 1 1 1, 1 1 1 1, 1, 14, 7, − = ⇒ = + =, −, =, =, v u f, v f u 30 750 750 375, Sol 24: Image from L1:, 375, ⇒ v = cm,, 7, L1, v 375 / 7, 1, ⇒m=, −, Magnification, m = =, 20 cm, −750, u, 14, 3.0 cm, ⇒ diameter of imageof bulb,d1 =, =, 0.214 cm, P1, 14, 20 cm, , L2, I2, P2, , M, , R, , I1, , C, 40 cm, , X

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P hysi cs | 16.105, , Add (iii) and (iv), , 1 1 1 1, 1, 4 −3 1, = + =, −, =, =, v1 f u 15 20, 60, 60, , 3 4, 3, − −1 +, 1, 1, 2, 3, 2, +, =, v 6 30, 30, , ⇒ v1 =, 60 cm from L1, For lens L2:, u2 =, +40 cm,, , ⇒, , 1, 1, 1, 8+3, =+, =, v 2 15 40 120, , from lens (right from lens), , So final image is 90 cm right of lens., , 120, ⇒ v2 =, +, cm from L2, 11, This final image should lie at the centre of curvature of, convex mirror, so MC = R = 5 cm, So,P2M= P2C2 − MC=, ⇒, =, x, , 1, 2, 1, −1, =−, = ⇒ v6 =, −90 cm, v 6 90 30 90, , 120, 120 − 55, − 5=, cm, 11, 11, , Exercise 2, Single Correct Choice Type, , 65, = 5.91cm, cm, 11, , Sol 1: (B) Distance of image due to plane mirror from, object will be 60 cm. So OO’ =60 cm. So distance of, image from convex mirror PO’ =[OO’-OP]=10 cm, , Sol 25: For equi-convex lens radius R= f= 30 cm, Refraction at surface I: Air to glass, 3, −1, 3/2, 1, 2,  ..............(i), −, =, v1, ( −90) +30 cm, , M, , ... (i), , O, , O’, 30 cm, , Refraction at surface II: Glass to water, , 4 3, −, 4/3 3/2, 3 2  ..............(ii), −, =, v2, v1, −30 cm, , ... (ii), , Add (i) and (ii), , 20 cm, , M’, , 1 1 1, + = will give focal length of, v u f, mirror by putting u=-50 cm, v=+10 cm,, , ⇒Mirror formula, , So,, , 3, 4 3, −1 − +, 4, 1, 2, 3, 2, +, =, 3v 2 90, 30, , P, , 1 1, 1, 4, 2, 25, =, − =, =, ⇒ f=, f 10 50 50 25, 2, , Radius of curvature R=2f =25 cm, , 4, 2, 1, ⇒, =−, ⇒ v2 =, +120 cm, 3v 2 90 90, , near end, , For mirror, image of lens acts as object., , F, , For mirror u3=+ (120-80) = 40 cm (right from mirror), So v3 = -40 cm (left from mirror), Refraction from surface II after reflection from mirror., u4=-40 cm (left from surface II), 3 4, −, 3/2 4/3 2 3 , −, = .........(iii), −40, +30, v5, Refraction at surface I: Glass to air, 3, 1−, 1 3/2, 2 .........(iv), −, =, , v6, v5, −30, , ... (iii), , ... (iv), , Sol 2: (A) Mirror formula, 1 1 1, 1 1 1 u−f, fu, + = ⇒ = − =, ⇒v=, ., v u f, v f u, fu, u−f, , At near end |u|> |f|. u and f both are negative ., So v is negative. At far end we have u = −∞ . So mirror, 1, 1 1, 1, formula gives, = −, = ⇒ v ∞ =f ., v ∞ f −∞ f, For near point |vnear| > |f|.

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1 6 . 1 0 6 | Geometrical Optics, , Image length, , | v near | − | v ∞=|, ⇒ ∆I =, , Sol 6: (D) Lens formula, , fu, fu − uf + f 2, −=, f, u−f, u−f, , 1 1 1, − =; u = -30 cm,, v u f, , f = 20 cm, , 2, , f, u−f, 2 cm, , Sol 3: (C) Velocity of image, , P, h1, ,  v  du dv,  v , dv, =, −  ;, =, −,  .4 cm s−1 ., 2, 2 , , , , dt, dt, dt, u, 20,  , , , 2, , 2, , h2, , From mirror formula, , 1, 1, 1, 3−2, 1, =, −, =, =, ⇒ v = 60, v 20 30, 60, 60, , 1 1 1, 1, 1, 1, 1, 1, 3 −5, 1, =− ⇒ = −, = −, =, =, −, v f u, v −12 −20 20 12, 60, 30, , ⇒v=, −30 cm ⇒, , 0.5 cm, , axis, , v, 60, Magnification m = ⇒ m =, =−2 ⇒ h2 =−2h1, u, −30, ,  900 , dv, −1, =, −, −9 cm s−1,  4 cm s =, dt,  400 , , So h2 =−2 × 0.5cm =−1cm (below axis), So Image of P is 1.5 cm below XY., , d2, Sol 4: (D) Area of mirror, A1 = π, ., 4, , Sol 7: (B) Distance between object and screen is D,, displacement of lens is d, and so focal length of lens is, , Area left after putting opaque,, , d2, d2, d2, A2 = A1 − π d2 = π d2 − π d2, A2 = A1 − π 16 = π 4 − π 16, 16, 4, 16, d2 , 1  3 πd22 3, ⇒ A2 =, π d2  1 − 1  =, 3 πd =, 3 A1, ⇒ A2 =, π 4  1 − 4  =, 4 4 =, 4 A1, 4 , 4 4 4, 4, Focal length will not change and intensity become, , =, f, , D2 − d2 902 − 202, =, = 21.4 cm, 4D, 4 × 90, , Sol 8: (A) Object size, , 3, I., 4, , Sol 5: (B) Rays should fall normally on plane mirror., This will happen if rays become parallel to principal axis, after passing through lens. So OL = f=30 cm., , O=, , I1 × I2 =, , 6 cm× 3 cm, , O = 4.24 cm, , Sol 9: (A), , 1 1 1, + =, v u f, , Let u = + x (virtual) and |f| = -f (concave mirror), , ⇒, , |f|x, 1, 1 1 −x − | f |, =−, − =, ⇒v=, → ( −)ve, v, |f| x, |f|x, −x − | f |, , So v is always negative when u is positive (+x)., , L, , 15 cm, , Multiple Correct Choice Type, Sol 10: (B, D) Slope of reflecting surface at the desired, point will be tan 45 = 1,  πx , dy, πx π, L, = 2cos   = 1 ⇒, = ⇒x=, dx, L, 3, 3,  L ,  L  2L, π, ⇒ y=, sin  =, , , 3 π, 3, , 3L, π

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P hysi cs | 16.107, , Sol 11: (B, C) Length of object is AB = (50-20) cm, =30 cm. After first reflection from plane mirror AB is, inverted to B1A1 with distance from convex mirror as, shown in figure., Image of A in convex mirror, , So as µp increases , δm also increases., v1, , 60 cm, u1, B1, , δ +A, sin  m, , µ, δ,  2 , If m is minimum deviation, p =, µs, A, sin  , 2, , A1, , A’, u2 = 90cm, , B’, v2, , 1 1 1, 1, 1, 1, 1, =−, ⇒, = −, = ⇒ v1 =, +30cm, v1 f u1, v1 60 −60 30, , Sol 15: (B, C) The minimum length of a plane mirror to, H, see one’s full height in it is , where H is the height of, 2, man. The mirror can be placed anywhere between the, centre line BF (of AC) and DG (of CE). Eye is at C., A, x, x, , Image of B in convex mirror, , F, , B, C, , 1 1 1, 1, 1, 3+2 1, = −, = −, =, =, v 2 f u2 60 −90 180 36, , y, , ⇒ v2 =, +36 cm ⇒ A'B' =, (36 − 30)cm =, 6 cm, , y, , 1, , (x + y), , D, E, Man, , th, , Second image A’B’ is virtual and   of magnification, 5, w. r. t. AB and erect., 1, , 2, , i, , P, , i, , G, , Sol 16: (B, D) The distance PQ1 and PQ2 will not change, as the mirror MM’ moves with speed v perpendicular, to its length., S, , P, , Q1, , Wall, , Q2, , i’, , i’, O, , Now formula for speed of image for convex mirror, dv v 2 du, =, is,, . As object moves towards mirror, the, dt u2 dt, image also moves towards the mirror., Sol 12: (B, C) i + i’ =90o from figure, At point P, µ1 sin i > µ2 , , … (i), , At Q,, , … (ii), , µ1 sini' > µ3 or µ1 cosi > µ3 , , Squaring and adding (1) and (2) to get, , µ12, , >, , µ22, , + µ32, , ⇒, , µ12, , − µ22, , >, , µ32, , ⇒, , µ12, , − µ32, , >, , µ22, , Sol 13: (A, C) Angle of deviation δ= ( µ -1)A., Sol 14: (A, B, D) Angle of deviation δ= (i1 + i2 ) − A . i1, is angle of incidence and i2 is angle of emergence and, angle of incidence and emergence are interchangeable., , V, M’, , M, , Sol 17: (A, C) µ2 > µ1 Rays from real object will be, deviated away from radius of curvature and hence, will becomes more diverging. For virtual object the, deviated rays may converge on the principle axis., µ1 > µ2 : For virtual object the deviated rays will, converge on principle axis., For real object the refracted ray will deviate towards, radius of curvature and may coverage on principle, axis., Sol 18: (A, D) The image formed by a convex mirror is, always, virtual and erect. So convex mirror cannot form, inverted image of OA. Option B and C are ruled out.

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1 6 . 1 0 8 | Geometrical Optics, , Sol 19: (A, C) u = -40 cm; f = +20 cm, , f =R−, , 1 1 1, 1 1 1 1, 1, 1, − = ⇒ = + =, −, =, v u f, v f u 20 40 40, , v, 40, ⇒ v =+40 cm, and magnification m = =, u −40, , R, sec60 = R − R = 0, 2, , o, = 180o − 2 × 60=, 60o . Ans (D), Sol 28: (D) Deviation, , ⇒ m =−1 ⇒ y and z coordinate’s values will change, their sign but magnitude will remain the same., , o, , 60, o, 60, , ⇒ y image =, −1 , zimage =, +1, , R, C, , Sol 20: (B, D) Convex mirror and concave lens for, diminished, virtual, correct image of object., Assertion Reasoning Type, Sol 21: (C) The wall does reflect light, but the reflection is, irregular. Reflected rays are deviated in different directions., , R, R, R − sec0 =, Sol 29: (B) f =, 2, 2, Sol 30: (B) Spherical aberration cannot be completely, eliminated, but it can be minimized by allowing either, paraxial or marginal rays to hit the mirror., , Sol 22: (A) This phenomena is called spherical, aberration. The rays close to the principal axis are, focused at the geometrical focus F of the mirror as, Sol 31: (D), given by mirror formula. The rays farthest from the, principal axis are focused at a point somewhat closer, (2n − 1)λD, =, ymin =, (n 1,2,...........), to mirror., 2d, Sol 23: (D) For object in liquid. dapparent =, , dactual, µ, , , 1, For a slab: normal shift ∆x= t  1 −  where t is, µ, , thickness of slab., Sol 24: (D) Two image will be formed one for each, mirror., Sol 25: (D) If a plane mirror is moved such that its, perpendicular distance from the point object does not, change, then the image will not move., , 5λD, For n =, +3, ymin =, 2d, , Sol 32: (A) Shift in the fringe due to the glass slab is, D, ∆y = (µ − 1)t where t is thickness of glass slab. Due to, d, glass slab path of ray from S2 gets increased by (µ − 1)t ., Sol 33: (A) Path of rays 1 is more than path of ray 2 by, a distance dsin α . Draw perpendicular S2M from S2 to, ray 1., , Comprehension Type, , S1, , Sol 26: (B) Paraxial rays are focused at the geometrical, focus F of the mirror. The marginal rays are focused at a, point F’ somewhat closer to the mirror., Sol 27: (D), , O, -40, , O, , P, , 1, S2, , y, , 2, I, , P, , , , X, , ∠MS2S1 =, α and MS1 =, d sin α, This path difference is suffered before passing the slits, S1 & S2. After passing through the slits, path of ray from

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P hysi cs | 16.109, , S2 is increased by (µ − 1)t . For net path difference to be, zero at point P we have,, S1, , M, , 1, , , , Sol 5: (B) lcutoff =, , Sol 6: (C) The refractive index n for meta-material is, negative., , d, , Hence, S2, , 2, , d sin α = (µ − 1)t, , sin θ1, , sin θ2, , is negative., , Thus if θ1 is negative, θ2 will be negative. So the current, choice is C., Sol 7: (B) N=, ,  (µ − 1)t , ⇒ α =sin−1 , ,  d , , Match the Column, Sol 34: A → p, r; B → q, s; C → p, q, r, s; D → p, q, r, s, , Sol 1: (C, D), , 1, 1, 1, =, + (mirror formula), f, v, u, , F = – 24 cm, , n0 / 8 3, sin=, θc, =, n0 / 6 4, , Now applying Snell’s law in region I and region III, , or, , ∴, , n0, 6, , sin θC, , 1, 13 1, sin, =, θ, sin=, θC, =, , 6, 64 8, 1, θ =sin−1  , 8, , Sol 3: (A, C, D) In case of concave mirror or convex lens, image can be real, virtual, diminished, magnified or of, same size., (B) In case of convex mirror image is always virtual (for, real object)., Sol 4: (A) At minimum deviation ( δ = δm ) :, r1= r2=, , c, which is choice B, n, , =, λair ×, , vm, , v air, , =, λair ×, , vm, , nair, , n, , c, v air, , c,  v = , n, , , nm, , λair, , c, , ×, , (, , and nair 1, =,  nm n=, , ), , So choice D is wrong, , Sol 2: (B) Critical angle from region III to region IV, , n0 sin, =, θ, , c, = v=, v, , v, Also frequency ν = since v remains unchanged, λ, v air v m, =, λair λm, , ⇒ λm = λair ×, , Previous Years’ Questions, , hc, (independent of atomic number), eV, , A 60o, =, = 30o (For both colours), 2, 2, ,  1, , – 0  + (1.2 – 1), Sol 8: (B) PT = (1.5 – 1) , 14, , , =, , , 1 , 0 –, , –14, , , , 0.5, 0.2, 1, +, =, 14, 14, 20, , f = + 20 cm, , 1, 1, 1, –, =, v, –40, 20, 1, 1, 1, 1, =, –, =, v, 20, 40, 40, \ v = 40 cm, Sol 9: (B) Object is placed at distance 2f from the lens., So first image I1 will be formed at distance 2f on other, side. This image I1 will behave like a virtual object for, mirror. The second image I2 will be formed at distance, 20 cm in front of the mirror, or at distance 10 cm to the, left hand side of the lens.

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1 6 . 1 1 0 | Geometrical Optics, , Now applying lens formula, , Hence, l will depend on n (z) but not on n2., , 10 cm, I3, O, I2, , 1, , n1 = constant, , I1, , 1, , 6 cm, , n(z), 30 cm, , 10 cm, , 20 cm, , z, , 1 1 1, − =, v u f, 1, 1, 1, −, =, v +10 +15, , ∴, , or, , d, , d, l, , n2 = constant, , v = 6 cm, , , , 2, , Therefore, the final image is at distance 16 cm from the, mirror. But, this image will be real., This is because ray of light is travelling from right to, left., Sol 10: (3) For v1 =, v2 =, , Taking the mirror to be straight, the image I1 after, reflection will be formed at 50 cm to the left of the, mirror., , 50, m, u1 = – 25 m, 7, , 25, m, u2 = – 50 m, 3, , Speed of object =, , On rotation of mirror by 30° the final image is I3., So x = 50 – 50 cos 60° =25 cm., , 25, 18, ×, = 3 km/h., 30, 5, , Sol 11: (B), 1, For the combination, =, feq, , Sol 13: (A) First Image I1 from the lens will be formed, at 75 cm to the right of the lens., , and y = 50 sin 60° =25 3 cm., I3, , ( µ1 − 1) + ( µ2 − 1), R, , 50 cm, , R, , o, , 30, , feq = 20, , o, , 30, (0, 0), , I2, , Here u = - 40, f = 20, , 50 cm, , v = 40, Sol 12: (A, C, D) From Snell’s Law, n1 sin, =, θi n ( d) sin=, θd n2 sin θf, , The deviation of ray in the slab will depend on n (z), , Sol 14: (A) Let angle between the directions of incident, ray and reflected ray be θ, cos=, θ, , (, , 1 ˆ, i + 3 ˆj, 2, , ), , , cos=, θ, , (, , ) (, , 1 ˆ, 1, i + 3 ˆj . ˆi + 3 ˆj, 2, 2, , )

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1 6 . 1 1 2 | Geometrical Optics, , 1, 10, 1, f, , , , Sol 23: (A, C) θ ≥ c, , 1 , ,  1 R 2  ⇒ f 35 cm, =, 2,  1.5,  1, 1 , − 1  , −, , , , 7/6,   R1 R 2 , , (1.5 − 1)  R1, , −, , ⇒ 90° − r ≥ c, , ⇒ sin ( 90° − r ) ≥ c, ⇒ cos r ≥ sin c, , n, sin i n1, =, and sin c = 2, sinr nm, n1, , So in second medium, final image is formed at 140 cm to, the right of the lens. Second medium does not change, , Using, , Mm M, M, 2, 2, the magnification by mirror.=, So 2, =, 7, M1, Mm M, , We get, sin2 im =, , 1, , 1, , n12 − n22, 2, nm, , Putting values, we get, correct options as A & C, , Sol 21: (B) For Ist refraction, , n2, , 1 1.5 1 − 1.5, −, =, −10, v −50, , nm, , r, , , , n1, , i, , ⇒v=, 50 cm, , For IInd refraction, Sol 24: (A) i = β + θ, , 1.5 1 1.5 − 1, −, =, ∞ −x, +10, , For α= 45° ; by Snell’s law,, , ⇒x=, 20 cm, , P, , ⇒d=, 70 cm, , , , Sol 22: Snell’s Law on 1st surface:, sin r1 =, , 3, , 2n, , 3, = n sin r1, 2, , , , i, , …(i), , n =2, Q, , 4n2 − 3, ⇒ cos r1 =1 −, =, 2n, 4n2, 3, , r1 + r2 = 60° , , , , 1 × sin=, 45°, …. (ii), , R, , 2 sin β, , ⇒ β= 30°, , Snell’s Law on 2nd surface:, , For TIR on face PR,, , n sin =, r2 sin θ, Using equation (i) and (ii), ,  1 , β + θ = θc = sin−1 ,  = 45°,  2, , nsin ( 60° − r1=, ) sin θ, , ⇒ θ= 45° − β= 15° ., ,  3, , 1, n, cos r1 − sin r1  =, sin θ, 2, ,  2, , Sol 25: (A, D) For refraction through lens,, , dθ, d  3 , 2, , ,  4n − 3 − 1  = cos θ,  , dn  4 , dn, , For θ= 60° and n = 3, , ⇒, , dθ, =, 2, dn, , 1, 1, 1, v, −, =, and −2 =, u, v −30 f, ∴ v =−2u =60 cm, ∴f =+20 cm

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P hysi cs | 16.113, , For reflection, , 1, 1, 2, +, = ⇒ R = 30 cm, 10 −30 R, , 1 1, =, (n − 1)  R1  =, f 20,  ,  , , 5, ∴n =, 2, , The faint image is erect and virtual.