Consider a ray of light incident from air

Question:

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)

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

Solution:

The correct answer is (a)

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

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