Question

A ray of light is incident on the surface of a glass plate of thickness $t$. If the angle of incidence $\theta$ is small, the emerging ray would be displaced side ways by an amount : [take $n=$ refractive index of glass].

• A. $t\theta n/(n+1)$
• B. $t\theta (n-1)/n$
• C. $t\theta n/(n-1)$
• D. $t\theta (n+1)/n$

Answer 1

Answer: (B)

$t\theta (n-1)/n$

We have, Lateral shift $=\frac{t\sin (i-r)}{\cos r}$ and $\frac{\sin i}{\sin r}=n$

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

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

Since i is very small, r is also very small.

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

Here $i=\theta$

$r=\frac{\theta }{n}$

Since i and r are small,

$\sin i\Rightarrow i$

$\sin r\Rightarrow r$

$\cos i\Rightarrow 1$

$\cos r\Rightarrow 1$

Lateral shift $=t(\theta -\frac{1\times \frac{\theta }{n}}{1})$

$=t\theta \left(\frac{n-1}{n}\right)$