Page 1 :
~ Example 15.1: A soap film 5 x 10 ~° cm thick is viewed at an angle of 35° to the normal,, Find the wavelengths of light in the visible spectrum which wiil be absent from the reflected light, (u= 1.33)., , Solution: Let ibe the angle of incidence and r be the angle of refraction. Then, , sin i sin 35°, L=— “. 1.33 =—— “. r= 25.5° and cos r = 0.90, sin r sin r, , , , ‘Scanned with CamScanner

Page 2 :
The condition for destructive interference is 2 ¢ cos r = mh., , Using different values for m in the above relation, we get following values for wavelengths., Whenm=1, A,=2x1.33x5x10-5cm x 0.90 = 12.0 x 10~* cm =120pm., , A= (2X 1.33 X5x 10-5 em x 0.90) +2= 6.0 x 107% cm= 6000 A., , When m = 2,, When m=3, A,= (2x 1.33 x5x 10-5 om x 0.90) +3 = 4.0 x 10 ~* em= 4000 A., Whenm=4, Ay=(2X 1.33 x 5x 10-5 em x 0.90) +4 = 3.0 x 10~> om= 3000 A., , Out of the above wavelengths, 2, = 6000 A, and A, =4000 A lie in the visible region. Therefore,, these two wavelengths are absent in the reflected light., , Example 15:2: A glass wedge of angle 0.01 radian, wavelength 6000 A falling normally on it. At what distanc, fringe be observed by reflected light? :, , Solution: Given that @ = 0.01 rad, m = 10, A= 6000 x 10 -8 cm., , The condition for dark fringe is 2 =m, , is illuminated by monochromatic light of, e from the edge of the wedge will the 10", , The angle of the wedge 09= us or t=Ox, , x, 20x=maA, mk 10x 6000x107 ®cm, , or x=— = =3, 20 2x0.01, , Example 15.3: A beam of monochromatic light of wavelength 5.82 x 10-7 m falls normally, onaglass wedge with the wedge angle of 20 seconds of an arc. If the refractive index of glass is 1.5,, , find the number of dark fringes per cm of the wedge length., , Solution: Given wedge angle 6 = 20" = TLL radians, A= 5.82 x 10-7m, p= 1.5., 60 x 60 x 180, , X 5.82107 ’mx 60x 60 x180, 2 eee = 2 mont, , Fringe width B= 500 Ix15X20XT, , 1, , Number of fringes per em = 9 5 cm = 5 per cm., » Example 15.4: A thin equiconvex lens of focal length 4 m and refractive index 1.50 rests on, and in contact with an optical flat, and using light of wavelength 5460A, Newton’s rings are viewed, tomnally by reflection, What is the diameter of the 5" bright ring?, , Solution: Given that m=5, 4 = 5460A = 5460 x 10-1 m, f= 4m, =15, , R, , f, jf, , mii A, 4am | R orR=4m., , The ai, * ameter ofthe m' bright ring is given by, 07 mx 4m = 6.2 mm, D,, =.[2(2m-1) AR = 2(2x5-1)x5460%1, , ‘Scanned with CamScanner, , 1 Ls f, We know that == rf -4| . Here R, = Rand R,=- R

Page 3 :
Example 15.5: Newton’s rings are observed in reflected light of A = 5.9 x 19-5., , "i &m,, diameter of the 10! dark ring is 0.5 om. Find the radius of curvature of the lens and the thigys, te, , e, the air film. 88 0p, Solution: Given that 1. = 5.9 x 10~> cm, m = 10., The radius of m‘ dark ring is given by, r (0.5cm)?, , ee “md 1059x107 Sem, The thickness of air film is given by, md — 10x5.9x10° om, 2o. 2 ‘, Example 15.6: In a Newton’s rings experiment, the diameter of lo dark ring diary, wavelength 6000 A in air is 0.5 cm. Find the radius of curvature of the lens., , (D/2) _ (0.5x10 /2)?m?, , ion? i R=——— = —\_ ——_ = 1.04m, Solution: Radius of curvature, mh 10x6000x10" m, , Example 15.7: In a Newton’s rings experiment the diameter of the 15" ring was found tobe, 0.59 cm and that of the 5'* ring was 0.336 cm. If the radius of the plano-convex lens is 10, calculate the wavelength of light used., De ,-Dz De —D2 (5.9-3.36)* x10 m2, Solution: 4 =—“*? st = 5880 A., 4pR 4xl0xR 4x10xlm, , Example 15.8: In a Newton’s rings experiment the diameter of 10'* ring changes from 1.40, , to 1.27 cm when a drop of liquid is introduced between the lens and the glass plate. Calculate the, refractive index of the liquid., , (Pn), (Pn), Example 15.9: In a Michelson interferometer 200 fringes cross the field of view when the, movable mirror is moved through 0.0589 mm. Calculate the wavelength of light used., . 2d _2x0.0589x10?m, Solution: A=— =, m 200, , _. Example 15.10: In an experiment for determining the refractive index of gas using ae, interferometer a shift of 140 fringes is observed, when all the gas is removed from the tube. If, , wavelength of light used is 5460 A and the length of the tube is 20 cm, calculate the refractive index, of the gas. F, , , , = 106 cm=1.06m, , , , ‘ t =2.95.m., , 0 cm,, , , , (1.40 cm)?, (1.27 em)’, , air, , , , , , Solution: H= =1.215, , lig., , , , =5890A, , Solution: Given that = 140, 1 = 5460 A = 5460 x 10-"m, 7=20 em = 0.2 m., , _, (ar 140 x5460x107° m, w=te() = 1+ SxO Sia = 1.00019., , , , Example 15.11: A glass microscope lens (t= 1.50 is coated with magnesium fluort, : p HL ) 00 A)- wh, = 1,38) film to increase the transmission of normally incident yellow light (A = 58, minimum film thickness should be deposited on the lens? rt ;, , Solution: Given that 1, = 1.50, w,= 1.38, 1= 5800 A., , Minimum film thickness twin pete SRD, 4u, 4x1.38, , , , A =1050A, , , , Scanned with CamScanner