Page 1 : Ray OPTIcs AND Ormica, j, , , , , , Where 1 = angle of incidence, t= thickness of parallel slab, Mt = refractive index of medium, , Actual depth, 26. Refractive index of medium = Apparent depth, 27. Increase of about 4 min,, , 28. The peak (specific valuc of the incidence angle for a, hight ray moving from a denser medium to a rarer], medium, for which corresponding angle of refraction is, 0° is called the critical angle,, , 29. If 4 is the refractive index of denser medium 1 with, Tespect to rarer medium 2 then, , 30. approx 244°, 31. approx 42°, , 32. Itis used in optical instruments like periscope, photometer], ete, , 33. Optical fibre is based on the phenomenon of total internal, reflection of light., , 34. Total internal reflection, , , , fovu, , 38. If focal length is in metre then unit of power of lens is, dioptre (D)., , 39. The power of lens for convex lens is positive and power}, , of lens for concave lens is negative., |@ Effective power of lens =P, +P,+ Pat,, , , , , , , , , , = a ——————————, , Ex.20. Find the position of the image formed by, , combination of lenses in the given diagram ;, NCERT/Examplei919), , S=+10,-10 +30cm, , , , , , , , Sm 1) —4, cm cm, , Fig}, , , , , , Sol., , Ex.21, A rectangular piece of glass is cut fr, , Sol., , Ny, Image formed by first lens 8,, 1 ot 1, wim A tay Orr, yin A A, =+, where u; =-30 cm, f 10 cm «, Bt, UW 30° 10 / «, > vy, =+15 cm, is image i it behaves like y:, This image is real but it bel haves like Virtua oo, second lens. ie eeas = :, Therefore u2= 15- cm ;, Image formed by second lens :, ait, v2 te iol. F, where u= 10cm ;, f,=- 10cm :, then Wao, _, This virtual image is formed at infinite dig, ‘, left:side of second Jens. It works 8S objec, ' 4, . Tens: 7 ad ty, vad, 3% f, where U3 = 00, f; = +30 cm, ie A, v3 9 30, > v= +30 cm, , Final image is formed at 30 cm distance, , towards fh, of third lens i.e. at its focus. 4, , Eve ¢ ‘OM one Sie, The radius of this circular plane is 19 cm, Th, , plane is raised in the middle, An object is Pacey, , at 30 cm distance in front of Plane in the ain, Where will the:image of this object be forme EEE, If refractive index of glass is 1.5, 2x23,, According to formula, , , , Sol., Given that w= 1.5, 1 =~30 em, R = 10 cm (come, , Surface), , , , LS, 1 os, 15,1 _ 0s Bx.24,, 7 V0 =

Page 2 : will be the value of, , , , Ray OrTics AND Orticat INSTRUMENTS, 15 1 1 Fo i, or Y 7 20 30 incidence angle?, , Bi], , a, nN, , 1, or yO 60, v= 60* 15=90em, gx.22. The focal length of a convex lens is 12 cm. Radius, of curvature of spherical surfaces of the lens are, 10 cm and 15 cm respectively. Find refractive, index of material of the lens., , , , 1 1 7, Sol. Formula : Tf =(u- lg -<, , Here f= 12 cm, R, = 10 om, R,=-15 em, By placing these values in the formula :, , 1, = (u-Dx{ S-——, a7 > +), or, , 1 342, or a os 2), , , , 5, = (p-l)x 3), eS a, 13, =i4+-=>, mea, w= 15, , Miscellaneous Examples, , 3x.23. Velocity of light in air is 3 * 10° m/s. If refractive, index of glass is 1.5 then find velocity of light:in, the glass?, , c, , ¥ c, iol, Formula #=— or ¥=, Vv u, , given that c=3 x 10* m/s and p= 1.5, By placing values in the formula :, , _ 3x08, , y =2x10° mis., , to each other then wha', , angle of incidence and angle, , In Diagram, i and rare, ' Reflected ray and refracted, , of refraction respectively., ray arc mutually perpen, , Sol., , dicular to each other, , , , , , , , |g 9.42:, , According to diagram, .j+.90° + r= 180°, , or i+ r= 90°, , or “p= (90°- i), , Now, according the snell’s law :, , _ sini, He Snr, sini sini :, = — = lan, sin(90°-i) cost, [: sin(90°-1) = cosi], Now 1.62 = tani, , or i= tan!.(1.62) = 58°19", , Ex.25. Absolute refractive index of glass and water are, 3 4, 5 and = respectively. Find out ratio of speed of, , light in glass and water., , ¢, Sol. From Formula : #=~,, , or, , , , Ex.26. A person standing on the bank of a pond, when, looking down vertically, a fish appears 2m below, , 1x.24. A ray of light incidents at rectangular block of, glass having refractive index 1.62. If reflected, and refracted rays are mutually perpendicular, , the water surface. If the refractive index of water, is 1.33 then what will be the actual depth of the

Page 3 : |, , feh®, ty, Ket Feruls refimetion indew, tinen thar po 1 he 2m, hs, then tae?, = 264m, , h $33, Fa? A rectanptiler piece of glace 3 cm thick is placed, fn rp paper, Hom mans centimetres ell! the, letters written on the paper appear above?, , j ', , Kol Formula :i, Gren that re Sem. pe tS, By plactny the values in the formala, a 1, det sAry-=], 8 3 cm, , fa.28. Speed of fight in medium 1 is 2 * 10° m/s and, speed of fight in medium 2 is 2.4 * 10° m/s. If, ligitt ray ts moving from medium I to medium 2, , then find value of critical angle., , ’, 1, , , , Sel, Formula Ma, Given thats, = 2 © 10° m/s and v, = 2.4 » 10" m/s, 2510 1, , a., He” Santo 12, or wri?, es clear that medium | 1s denser medium and medium, 2 © rarer medium, 1, a0, A sant, C | I $, wn on ” Ly 12 ©, {5 }, Lor = sin’ | 6), , £4.29, Tie focul length of a conves lens of glass (y=, 1,8) is 2 eat Hf this lens iy immersed in a liquid,, what will be its focal length? If the refractive, , wdes of liquid is 1.25,, , Sel. Given that A, 1S, w= 128, f =42 em, eh, 18 6 ‘, As” ay iss, , 430.4 light, Fx.30.¢ path of the converged beam, , (i), (ii), , Sol., (i), , (ii), , Sol,, , nay OFTICS AND Orne, Su,, , , , N, From formula %, fs 2, ta, or, , , , beam converges at point p, , in the FR, of 12 cm from point P. At which gos, beam converge? If: |, , Lens is convex lens and its focay les, , « lens and its f; "i, Lens is a concave Ocal, ia 1 1 ol ‘ei, , Formulas fF > yn, , For convex lens f= +20 cm, u = +12 em, (object distance 1n the direction of light), , , , eat, 20 ~¥v 12, ls, or vy 20, 60 v= yz wt 75cm, , , , Therefore beam will converge at distay,, towards point P., , For concave lens f= -16 cm, , By placing the values in the formula :, , lo y 12, , . I, , or =, ¥, , v, , , , fs 48 cm, therefore, beam will converge at a distance !, towards point P. , Ex.3L. The focal length of a lens is 30 cm it!, object is placed at a distance of 10 cm!, first focus plane, How far will the ime, Object be formed on the other side of ™, , , , From Formula. x.x ee, given that x, =-1l0em, f =-30cm and/,, By Putting the values in above formula:

Page 4 : Ray Ortics AND OrricAL INSTRUMENTS, =10x, = -30?, , 230?, , =p = 90 cm, So that image distance from sccond focus = 90 cm, ~. Image distance from the lens = 30 + 90 = 120 cm, , Ex.32. A convex lens of focal length 30 cm is placed in, contact with a concave lens of focal length 20, cm. This system will behave like a converging, lens or a diverging lens?, , , , or x, , , , 1, , Sol. f, Given that /, = + 30 cm and f, = -20 cm, From above formula : ., , 1 1 Lod, , A, 77 eo ome., 2-3_ 1, , 60 60, f=-60 cm =- 0.60 m, , ‘ 1, Power of lensP = 7 260, , =-1.66D, Therefore, system will work as diverging lens, (concave lens), , Ex.33. A optician makes a lens by combining the two, lenses of power +1.25 D and -2D respectively,, then find power of newly formed lens., , Sol. As we know that P = P, + P, :, given that P, = 1.25 D, P, = -2.0D, By placing values P = 1.25 — 2.00 =-0.75 D, , zx.34. The refractive index of glass with respect to, water is 9/8. If the speed of light in water is 2.25, x 10° m/s then find the speed of light in glass., , Vw, , tol. Formula: #4, = Vp, 9, given that f= g, , v, = 2.25 x 10° mvs., vy _ 225x108, , YF ug 9/8, , , , 8 108, = 2.25 x —x10, 2.25 * 9%, , = 2.0 x 108 m/s, , x.35, Speed of light in glass ( = 1.5) is 2 * 10° m/s, then find out the speed of light: in air., , , , a. Formula: 4,, , aWyeex 10° nV/s, , sabe, , P= 15 x2% 108, , . = 3.0 x 10° m/s., , Ex.36. The image of an object placed in front of a convex, lens is formed 4.6 cm in height on the screen, placed on the other side. A second image of 1.7, cm in height is obtained when the lens is moved, towards the screen'then find height of object?, , where, , , , Sol. Formla OF yhxh, given that l= 4.6 cm, L, = 1.7 cm, , Putting the value in the formula :, , O= 46x17 =V782 = 2:80 cm, , Ex.37. The focal length of a convex lens is 75 cm then, find its power. ., Sol. Given that f= 75 cm, = 0.75 m, , 0.75, , Ex.38. A pin is placed at a distance of 10 cm in front of, a convex lens of focal length 20 cm and refractive, index of the substance is 1.5. The surface away, from the pin is silver coated radius of curvature, of this surface is 22 cm. Where will the final, image of pin be formed?, , 1_2,2, , -oe tte, Sol. Formula: fi R, , I 1 7, Power (P) = F075 = 1,33 Dioptre, , given that f,=—- 20 om, R,= -22 cm, Putting the values in above formula :, 1 -2,2_ -2l, ee, f~ 20°22 110, , again, , , , , , , , , , , , , , 1 1/1, b : ete,, Putting the value in formula fou v, , =21 -1,1, oop, 10 «10 v