Question

A ray of light is incident on a parallel slab of thickness $\left(t\right)$ and refractive index $\left(n\right)$. If the angle of incidence $\left(i\right)$ is small, then the lateral displacement of emergent ray will be

• A. $\frac{ti(n-1)}{n}$
• B. $\frac{ti}{n}$
• C. $ti\left(\frac{n}{n+1}\right)$
• D. $\frac{ti(n+1)}{n}$

Answer 1

The correct option is A $\frac{ti(n-1)}{n}$

Lateral displacement is given by,

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

If $i$ is very small , $r$ is also small, then $\sin i\to i,\sin r\to r$ and $\cos r\to 1$.

From snell's law,

$\frac{\sin i}{\sin r}=n$

This can be rewritten as $\frac{i}{r}=n$

Therefore, the expression for lateral displacement takes the form

$S=(1-\frac{1}{n})ti$

$\Rightarrow S=\left(\frac{n-1}{n}\right)ti$

Hence, option (a) is the correct answer.