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Numericals Based on Interference of Light

Problem 1. A film of thickness 0.6 µm of soap solution having refractive index 1.4 is formed. It is observed at an angle of incidence 4°. Determine that wave lengths of visible light, absent in the reflected light.

 

Hint: The condition for a particular wavelength, to be absent from the reflected light is

 

2 µt cos r = A., 2λ., 3 λ., ……

 

Let these wavelengths be λ1, λ2 λ3, …. . .

 

µ= sin i/ sin r    =>  r =sin-1 (sin 4° /1.4) = 27°20’

 

λ 1 = 2 X 1.4 X 0.6 X 10-6 X cos 27°20′ = 14923 A

2 λ 2 = 2µt cos r ==> λ 2 = 7461 A

Similarly         3 λ 3 = 2µt cos r ==> λ3 = 4974 A

 

The visible region is 4000 to 7000 A

So the absent wavelength will be 4974 A.

 

Problem 2. A biprism is placed at a distance of 5 em in front of a narrow slit, Illuminated b sodium light (λ =5 890 A) and the distance between the virtual sources is found to be 0.05 cm. Find they width of the fringes observed in eyepiece placed at a distance of 75 cm from the biprism.

 

Hint:                                       B = λ D/ d = 5890 x 10-10 x (75 + 5) x 10-2 / 0.05 X 10-2

= 9.424 A.

Problem 3. When a thin piece of glass 3.4 x 104 cm thick is placed in the path of one of the interfering beams in a biprism arrangements, it is found that the central bright fringe shifts through a distance equal to the width of four fringes. Find the refractive index of the piece of glass. Wavelength of light used is 5.46 x w-5 cm.

 

Hint:                                       (µ- 1) t = nλ

µ = 1.64

 

Problem 4. A surface of refractive index 1.5 is to be coated with a film of magnesium fluoride (µ= 1.38) to minimise the reflection. Calculate the minimum thickness of film for normal incidence of light of wavelength 5000 A.

 

Hint:                           t = λ/4µ= 5000/4 * 1.38 = 906 A

 

Problem 5. Light of wavelength 6000 A falls normally on a thin wedge shaped film of refractive index 1.4, forming fringes that are 2 mm apart. Find the angle of the wedge.

 

Hint:                                       β = λ/2θµ  => θ = λ/2µβ

θ = 1.07 x 10-4 radian

 

Problem 6. Interference fringes are produced by monochromatic light falling normally on a wedge shaped film of cellophane of refractive index 1.4. If angle of wedge is 20 seconds of an arc and the distance between successive fringes is 0.25 cm, calculate the wavelength of light.

 

Hint:                                        λ = 2 θ µβ

= 2 x 20 xπ x 1.4 x 0.25 / 60 X 60 X 180

λ= 6790 A

 

Problem 7. A planoconvex lens of radius 300 cm is placed on an optically flat glass plate and is illuminated by monochromatic light. The diameter of 811′ dark ring in the transmitted system is 0.72 cm.

Calculate the wavelength of light used.

 

Hint:                                        r2 = (2n – 1) λ R /  2

λ= 5760 A

 

Problem 8. In the Newton’s rings arrangement, the radius of curvature of the curved surface is 50 cm. The radii of the 9th and 16th rings are 0.18 cm and 0.2235 cm respectively. Calculate the wavelength.

Hint:                                       D9 = 2 x 0.18 = 0.36 cm

 

D16 = 2 x 0.2235 = 0.447 cm

R =50 cm, p = 7

Then                                                    λ = ((Dn+p)2  - (Dn)2) / 4pR = 5015 A

 

 

Problem 9. In a Newton’s rings experiment the diameter of the 12th ring changes from 1.50 to 1.35 cm liquid. when a liquid is introduced between the lens and the plate. Calculate the refractive index of the

 

Hint:                           For liquid medium D/= 4nµλR                                 …( i)

 

and for air medium                 D2 2 = 4nλR                                                    … (ii)

Then                                                    µ = (D2/ D1)2 =1.235

 

Problem 10. In Newton ‘s rings pattern the diameter of 61″ bright ring is 0.5 cm. The wavelength of the light used is 4590 A. Light enters the film with angle of incidence of 39°. Calculate the radius of curvature of lens.

 

Hint:                                        2µt cos r = (2n+1) λ/2 and t = r2/2R

Then

rn2 = (2n+1)λR/ 2µ cos r

dn2 = 2(2n+1) λR/ µ cos r

R = 193.66 cm.