Question

Consider a ray of light incident from air onto a slab 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. $\frac{4\pi d}{\lambda} \left(1-\frac{1}{n^{2}}sin^{2}\theta\right)^{{1}/{2}} + \pi$
• 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)

From ray diagram, path difference $(\Delta x)$
$\Delta x = AD + DC - AB$
on geometry to get,
$AD = DC = d \sec \,r$
$t = AC = 2\,AD \,\sin\,r = 2dt \tan \,r$
so $t = AC \,\sin\theta $
$= 2d\tan r \,sin\theta$
so $\Delta x = 2d\, \sec \,r - 2\,\tan\,r\,\sin \theta $
$ = \frac{2d}{cos\,r} (1-\sin\,r \,\sin\,\theta) $.....(i)
from snell's law
$\sin \theta=n \,\sin\, r $
so, $\cos\, r=\sqrt{1-\sin ^{2} r }=\sqrt{1-\frac{\sin ^{2} \theta}{n^{2}}}$
(i) gives, $\Delta x=\frac{2 d x}{\sqrt{n^2}-\sin ^{2} \theta}\left(1-\frac{\sin ^{2} \theta}{n} \theta\right)=\frac{2 d}{\sqrt{n^{2}-\sin ^{2} \theta}}\left(n-\sin ^{2} \theta\right)$
$\therefore $ Phase difference $=\frac{2 \pi}{\lambda} \Delta x+\pi$
[an additional $\pi$ due to reflection at $D$]
$\Delta \phi=\frac{4\pi d}{\lambda} \sqrt{1-\frac{1}{n^{2} \,\sin ^{2} \theta}}+\pi$