Question

A thin film of oil (n = 1.27) is located on smooth, wet pavement. When viewed from a direction perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no light at 548 nm. (a) What is the minimum thickness of the oil film? nm (b) Let m1 correspond to the order of the constructive interference and m2 to the order of destructive interference. Obtain a relationship between m1 and m2 that is consistent with the given data.

Answer 1

Answer:

Explanation:

The problem is based on interference in thin films

refractive index of water is more than given oil so there will be phase change of π at upper and lower layer of the film .

a )

for constructive interference , the condition is

2μt = nλ where t is thickness of layer , μ is refractive index , λ is wavelength and n is order of the fringe

Putting the values

2 x 1.27 t = n x 640

2 x 1.27 t =  640 ( for minimum thickness n = 1 )

t = 252 nm .

b )

2 x 1.27 t = m₁  λ₁

for destructive interference

2μt = (2m₂+1)λ₂/2

2 x 1.27 t =(2m₂+1)λ₂/2

m₁ λ₁  = (2m₂+1)λ₂/2

2m₁λ₁  = (2m₂+1)λ₂

2m₁ / (2m₂+1) = λ₂ / λ₁

2m₁ / (2m₂+1) = 548/ 640

2m₁ / (2m₂+1) = .85625

2m₁ = .85625 (2m₂+1)

This is the required relation between m₁ and m₂