Question

A coin lies at the bottom of a lake 2 m deep, at a horizontal distance 'x' from the viewer situated 1 m above the surface of a liquid as shown in the figure. Refractive index of water is $\mu =\sqrt{2}$. Find the value of x, if refracted ray enters the eye of the viewer at an angle of ${45}^{\circ }$ from the vertical.

• A. $(1+\frac{1}{\sqrt{2}})\text{m}$
• B. $(2+\frac{1}{\sqrt{3}})\text{m}$
• C. $(1+\frac{2}{\sqrt{3}})\text{m}$
• D. $(3+\frac{1}{\sqrt{3}})\text{m}$

Answer 1

The correct option is C $(1+\frac{2}{\sqrt{3}})\text{m}$

Applying Snell's law at liquid-air interface.
$r={45}^{\circ }$
${\mu }_a=1,{\mu }_l=\mu =\sqrt{2}$
$\Rightarrow \mu sini={\mu }_{a}sinr$
or, $\sqrt{2}sini=\left(1\right)sin{45}^{\circ }$
or $sini=\frac{1}{2}$
$\therefore i={30}^{\circ }$


From geometry,
$tan{45}^{\circ }=\frac{1}{P^{\prime }Q}$
or $P^{\prime }Q=1m$
and, $P^{\prime }Q=PM=1m$
Similarly from $\Delta QMC,$
$tani=\frac{CM}{QM}$
or, $tan{30}^{\circ }=\frac{CM}{2}$
$\frac{1}{\sqrt{3}}=\frac{CM}{2}\Rightarrow CM=\frac{2}{\sqrt{3}}$
$\Rightarrow x=PM+CM$
or, $x=(1+\frac{2}{\sqrt{3}})m$

Why this question? Tip: Apply Snell's law at interface and focus on the geometry of the ray diagram.