Question

A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface (see figure).

How fast is the light travelling in the liquid ?

• A. $ 1.8 \times 10^{8}\,m/s $
• B. $ 2.4 \times 10^{8}\,m/s $
• C. $ 3.0 \times 10^{8}\,m/s $
• D. $ 1.2 \times 10^{8}\,m/s $

Answer 1

Answer: (A)

Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is $90^{\circ}$.
As shown in figure, a light ray from the coin will not emerge out of liquid, if $i > C$.

Therefore, minimum radius $R$ corresponds to $i = C$. In $ΔSAB$,
$\frac{R}{h}=tan\,C$
or $R=h\,tan\,C$
or $R=\frac{h}{\sqrt{\mu^{2}-1}}$

Given, $R = 3\, cm$, $h = 4 \,cm$
Hence, $\frac{3}{4}=\frac{1}{\sqrt{\mu^{2}-1}}$
or $\mu^{2}=\frac{25}{9}$ or $\mu=\frac{5}{3}$
But $\mu=\frac{c}{v}$ or $v=\frac{c}{\mu}$
$=\frac{3 \times 10^{8}}{5/3}$
$=1.8\times10^{8}\,m/s$