Question

A ray of light is incident on a parallel slab of thickness $t$ and refractive index $\mu$. If the angle of incidence $\theta$ is small, then the lateral displacement of the incident ray will be equal to,

• A. $\frac{t\theta (\mu -1)}{\mu }$
• B. $\frac{t\theta }{\mu }$
• C. $\frac{\mu t\theta }{\mu -1}$
• D. None of these

Answer 1

The correct option is A $\frac{t\theta (\mu -1)}{\mu }$

The lateral displacement of the incident ray is,
$d=\frac{t}{\cos r}\sin (i-r)$

$=\frac{t}{\cos r}(\sin i\cos r-\cos i\sin r)$

$=t(\sin i-\cos i\frac{\sin r}{\cos r})$

As, $\mu =\frac{\sin i}{\sin r}\Rightarrow \sin r=\frac{\sin i}{\mu }$

$\therefore d=t(\sin i-\cos i\frac{\sin i}{\mu \cos r})$
$=t\sin i(1-\frac{\cos i}{\mu \cos r})$
$=ti\left(\frac{\sin i}{i}\right)(1-\frac{\cos i}{\mu \cos r})$

As the angle $i$ is small, $\Rightarrow$ the angle $r$ will be even smaller.
$\Rightarrow \frac{\sin i}{i}\approx 1,\cos i\approx 1\text{and}\cos r\approx 1$
$$\text{As},\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1,\underset{\theta \to 0}{lim}\cos \theta =1$$
$\therefore d=ti(1-\frac{1}{\mu })$

$=t\theta \left(\frac{\mu -1}{\mu }\right)(\because i=\theta )$

Hence, $\left(A\right)$ is the correct answer.