Question

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

$(d=3mm,\theta ={45}^0$ and $\frac{n_{glass}}{n_{air}}=\frac{3}{2})$

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

Answer 1

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

We use the formula for redfractive index,

$\mu =\frac{sini}{sinr}$

This, gives, sin r = $\frac{\frac{1}{\sqrt{2}}}{\frac{3}{2}}$

Thus, $sinr=\frac{2}{3\sqrt{2}}$

Thus, $cosr=\frac{\sqrt{7}}{3}$

Now, applying simple geometry arguments,

$\frac{ABcosr}{ABsin{45}^0}=\frac{d^{\prime }}{d}$ we get

$d^{\prime }=\sqrt{2}dcosr$

Hence, d' = $\sqrt{14}$