Answer:
Explanation:
a , b )
The problem is based on interference in thin films
formula for constructive interference
2μ t = ( 2n+ 1 ) λ / 2 , μ is refractive index of layer, t is thickness and λ is wavelength of light.
n is called the order of fringe . If we place n= 0 , 1 , 2 etc , the thickness also changes . So constructive interference is possible at more than one thickness .
Put the value of λ = 116 nm . μ = 1.28 , t = 116 nm in the given equation
2 x 1.28 x 116 x 2 = ( 2n+ 1 ) λ
593.92 = ( 2n+ 1 ) λ
when n = 0
λ = 593.92 nm .
This falls in visible range .
c )
2μ t = ( 2n+ 1 ) λ / 2
Put λ = 593.92 nm , n = 1
2 x 1.28 t₁ = 3 x 593.92 / 2
t₁ = 348 nm .
Put n = 2
2 x 1.28 t₂ = 5 x 593.92 / 2
t₂ = 580 nm .