Question

For very small angle of incidence of a ray of light striking a glass slab is shown in the figure below

Which of the following expression will give the approximate result for lateral displacement? Use for small $\theta ,\sin \theta \approx \theta$ and $\cos \theta \approx 1$

• A. $d=ti(1-\mu )$
• B. $d=ti(1-\frac{1}{\mu })$
• C. $d=ti(\mu -1)$
• D. $d=ti$

Answer 1

The correct option is B $d=ti(1-\frac{1}{\mu })$

The formula for lateral displacement is,
$d=\frac{t\sin (i-r)}{\cos r}\dots \dots \left(i\right)$
From Snell's law:
$\frac{\sin i}{\sin r}=\mu \dots \dots \left(ii\right)$
for very small angle $i$, ${}^{\prime }{r}^{\prime }$ will be also very small.
$\Rightarrow \sin i=i,\sin (i-r)\approx i-r,\cos r\approx 1$
Thus from Eq. (i) and (ii),
$d=\frac{t(i-r)}{1}=t(i-r)$
$\Rightarrow d=ti[1-\frac{r}{i}]$
from Eq. (ii) $\frac{r}{i}=\frac{1}{\mu }$
$\Rightarrow d=ti[1-\frac{1}{\mu }]$