Question

Light of wavelength λ travels through a medium with an index of refraction n1before striking a thin film with an index of refraction n2 at an angle of incidence of 0. Some of the light is reflected off of and some is transmitted through the n2 thin film. The transmitted light travels a distance t through the thin film before encountering the n1 medium again. Some of the light reflects off of the n1 medium.What is the effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium, given that n1>n2?
A. t+λ/n2
B. 2t+λ/n2
C. 2t+λ/(2n2)
D. t+λ/(2n2)

Answer 1

Answer:

option C

Explanation:

The correct answer is option C

A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂  medium. Hence it travels 2 t distance more than the light that is reflected off n₂.

It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).  

Wavelength in n₂ is = $\dfrac{\lambda}{n_2}$  

Hence, path length difference = $2t +\dfrac{\lambda}{2 n_2}$