Question

A point source $S$ is placed at the bottom of a transparent block, made up of diamond, of height $12\, mm$. The block is immersed in a liquid of lower refractive index as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter $18\, mm$ on the top of the block. The refractive index of the liquid is _______(Take R.I. of diamond $=2.4$ )

Answer 1

As the light is passing from optically denser medium to rarer medium, $\theta$ is the critical angle.
From Pythagoras theorem,
$SB =\sqrt{ SO ^{2}+ OB ^{2}}$
$=\sqrt{12^{2}+9^{2}}=15\, mm$
$\therefore \sin \theta=\frac{ OB }{ SB }=\frac{9}{12}=\frac{3}{4}$
But, $\sin \theta=\frac{\mu_{2}}{\mu_{1}}=\frac{\mu}{2.4}$
$\therefore \frac{3}{4}=\frac{\mu}{2.4}$
$\Rightarrow \mu=1.8$