Question

A point source $S$ is placed at the bottom of a transparent block, made up of diamond, of height $12mm$. 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 $18mm$ on the top of the block. The refractive index of the liquid is _______(Take R.I. of diamond $=2.4$ )

Answer 1

Answer: 1.8

As the light is passing from optically denser medium to rarer medium, $\theta$ is the critical angle.
From Pythagoras theorem,
$SB=\sqrt{S{O}^2+O{B}^2}$
$=\sqrt{12^2+9^2}=15mm$
$\therefore \sin \theta =\frac{OB}{SB}=\frac{9}{12}=\frac{3}{4}$
But, $\sin \theta =\frac{{\mu }_2}{{\mu }_1}=\frac{\mu }{2.4}$
$\therefore \frac{3}{4}=\frac{\mu }{2.4}$
$\Rightarrow \mu =1.8$