Question

A ray of light is incident on a glass slab of width $3\text{cm}$ at point $\text{A}$. The lower portion of the glass slab is silvered. The lights get reflected from it and emerges out of the slab at point $\text{B}$. Then, the distance $\text{AB}$ is

• A. $\frac{3\sqrt{2}}{5}\text{cm}$
• B. $\frac{6\sqrt{2}}{7}\text{cm}$
• C. $\frac{3\sqrt{14}}{7}\text{cm}$
• D. $\frac{6\sqrt{14}}{7}\text{cm}$

Answer 1

The correct option is D $\frac{6\sqrt{14}}{7}\text{cm}$

Given:
$t=3\text{cm};\mu =3/2;i={45}^{\circ }$


From Snell's law,

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

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

$\Rightarrow \sin r=\frac{2}{3}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{3}....\left(1\right)$

In $\Delta ADC$
$\tan r=\frac{AD}{DC}$

$\Rightarrow AD=DC\times \tan r$

$\Rightarrow AD=3\times \frac{\sin r}{\sqrt{1-{\sin }^{2}r}}$

$\Rightarrow AD=\frac{3\times \frac{\sqrt{2}}{3}}{\sqrt{1-\frac{2}{9}}}$ [from Eq.(1)]

$\Rightarrow AD=\frac{3\sqrt{14}}{7}\text{cm}$

So, $AB=2AD=\frac{6\sqrt{14}}{7}\text{cm}$
[by symmetry]

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