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Q. A point source of light is kept at a depth of $h$ in water of refractive index $4/3$. The radius of the circle at the surface of water through which light emits is

WBJEEWBJEE 2008Ray Optics and Optical Instruments

Solution:

$\frac{\sin 90^{\circ}}{\sin C}=\mu$
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$\sin C=\frac{1}{\mu} \Rightarrow \frac{R}{\sqrt{R^{2}+h^{2}}}=\frac{3}{4}$
Squaring, $16 R^{2}=9 R^{2}+9 h^{2}$
$7 R^{2}=9 h^{2} \Rightarrow R=\frac{3}{\sqrt{7}} h$