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. $\frac{4\pi d}{\lambda }{\left(1-\frac{1}{n^2}si{n}^2\theta \right)}^{1/2}+\pi$
• B. $\frac{4\pi d}{\lambda }{\left(1-\frac{1}{n^2}si{n}^2\theta \right)}^{1/2}$
• C. $\frac{4\pi d}{\lambda }{\left(1-\frac{1}{n^2}si{n}^2\theta \right)}^{1/2}+\frac{\pi }{2}$
• D. $\frac{4\pi d}{\lambda }{\left(1-\frac{1}{n^2}si{n}^2\theta \right)}^{1/2}+2\pi$

Answer 1

Answer: (A)

From ray diagram, path difference $(\Delta x)$
$\Delta x=AD+DC-AB$
on geometry to get,
$AD=DC=d\sec r$
$t=AC=2AD\sin r=2dt\tan r$
so $t=AC\sin \theta$
$=2d\tan rsin\theta$
so $\Delta x=2d\sec r-2\tan r\sin \theta$
$=\frac{2d}{cosr}(1-\sin r\sin \theta )$.....(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, $\Delta x=\frac{2dx}{\sqrt{n^2}-{\sin }^2\theta }\left(1-\frac{{\sin }^2\theta }{n}\theta \right)=\frac{2d}{\sqrt{n^2-{\sin }^2\theta }}\left(n-{\sin }^2\theta \right)$
$\therefore$ Phase difference $=\frac{2\pi }{\lambda }\Delta x+\pi$
[an additional $\pi$ due to reflection at $D$]
$\Delta \varphi =\frac{4\pi d}{\lambda }\sqrt{1-\frac{1}{n^2{\sin }^2\theta }}+\pi$