Question

Consider a ray of light incident from air into a slap of glass (refractive index $n$) of width $d$, at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

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

Answer 1

Answer: (A)

In figure, a ray of light $A B$ is incident from air onto glass slab of width $d$ at angle $\theta$. It is reflected partially at $B$ and refracted at $B$ along $B C$ at $\angle r$. At $C$, the ray is partially reflected along $C D$ and partially refracted (not shown). To calculate phase difference between rays reflected from $B$ and $C_{r}$ we find

Time difference, $\Delta T=$ time taken to travel $B C$ in glass
$=\frac{B C}{v}=\frac{d / \cos r}{c / n}=\frac{n d}{c \cos r}$
From Snell's law, $n=\frac{\sin \theta}{\sin r}, \sin r=\frac{\sin \theta}{n}$
$\cos r=\sqrt{1-\sin ^{2} r}=\left(1-\frac{\sin ^{2} \theta}{n}\right)^{1 / 2}$
$\therefore \Delta T=\frac{n d}{c\left (1-\frac{\sin ^{2} \theta}{n^{2}}\right)^{1 / 2}}$
$=\frac{n \times T d}{\lambda}\left(1-\frac{\sin ^{2} \theta}{n^{2}}\right)^{-1 / 2}$
Phase difference $=\Delta \phi=\frac{2 \pi \Delta T}{T}$
$=\frac{2 \pi n d}{\lambda}\left(1-\frac{\sin ^{2} \theta}{n^{2}}\right)^{-1 / 2}$
As reflection at $C$ is from medium of higher refractive index, additional phase diff. of $\pi$ is introduced.
Hence required phase difference
$=\frac{2 \pi n d}{\lambda}\left(1-\frac{\sin ^{2} \theta}{n^{2}}\right)^{-1 / 2}+\pi$.