Question

A parallel beam of diameter $d$ is incident on the air-glass interface, as shown in figure. The diameter of the refracted light beam is

$(d=3\text{mm},\theta ={45}^{\circ }\text{and}\frac{n_{glass}}{n_{air}}=\frac{3}{2})$

• A. $\sqrt{12}\text{mm}$
• B. $\sqrt{14}\text{mm}$
• C. $\sqrt{6}\text{mm}$
• D. $4.5\text{mm}$

Answer 1

The correct option is B $\sqrt{14}\text{mm}$


We use the formula for refractive index,

$\mu =\frac{\sin i}{\sin r}=\frac{n_{glass}}{n_{air}}$

$\frac{\sin {45}^{\circ }}{\sin r}=\frac{3}{2}$

$\Rightarrow \sin r=\frac{\sqrt{2}}{3}$

$\therefore \cos r=\frac{\sqrt{7}}{3}$

Now, applying simple geometry arguments,

$\frac{AB\cos r}{AB\sin {45}^{\circ }}=\frac{d^{\prime }}{d}$

we get, $d^{\prime }=\sqrt{2}d\cos r$

$\therefore d^{\prime }=\sqrt{2}\times 3\times \frac{\sqrt{7}}{3}=\sqrt{14}\text{mm}$

Hence, option $\left(\text{B}\right)$ is correct.