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. $\frac{4\pi nd}{\lambda }{\left(1-\frac{1}{n^2}{\sin }^2\theta \right)}^{-1/2}+\pi$
• 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 $AB$ is incident from air onto glass slab of width $d$ at angle $\theta$. It is reflected partially at $B$ and refracted at $B$ along $BC$ at $\angle r$. At $C$, the ray is partially reflected along $CD$ 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 $BC$ in glass
$=\frac{BC}{v}=\frac{d/\cos r}{c/n}=\frac{nd}{c\cos r}$
From Snell's law, $n=\frac{\sin \theta }{\sin r},\sin r=\frac{\sin \theta }{n}$
$\cos r=\sqrt{1-{\sin }^{2}r}={\left(1-\frac{{\sin }^2\theta }{n}\right)}^{1/2}$
$\therefore \Delta T=\frac{nd}{c{\left(1-\frac{{\sin }^2\theta }{n^2}\right)}^{1/2}}$
$=\frac{n\times Td}{\lambda }{\left(1-\frac{{\sin }^2\theta }{n^2}\right)}^{-1/2}$
Phase difference $=\Delta \varphi =\frac{2\pi \Delta T}{T}$
$=\frac{2\pi nd}{\lambda }{\left(1-\frac{{\sin }^2\theta }{n^2}\right)}^{-1/2}$
As reflection at $C$ is from medium of higher refractive index, additional phase diff. of $\pi$ is introduced.
Hence required phase difference
$=\frac{2\pi nd}{\lambda }{\left(1-\frac{{\sin }^2\theta }{n^2}\right)}^{-1/2}+\pi$.