Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is (a) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\pi $ (b) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2} $ (c) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\frac{\pi}{2} $ (d) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+2 \pi $
Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is (a) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\pi $ (b) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2} $ (c) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\frac{\pi}{2} $ (d) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+2 \pi $

The correct answer is a) $ \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\pi $

Explanation:

When a slab of glass is put in air, the wave reflected from the upper surface (from a denser medium) suffers a rapid phase change of, but the wave reflected from the lower surface (from a rarer medium) does not.
An analogy between reflected light waves and the reflections of a transverse wave on a stretched string when it reaches a barrier is useful:
A beam reflecting from a material with a higher refractive index experiences a 180° phase change.