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Physics
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
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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
WBJEE
WBJEE 2008
Ray Optics and Optical Instruments
A
$ \frac{3}{\sqrt{7}}h $
38%
B
$ \frac{\sqrt{7}}{3}h $
19%
C
$ \frac{\sqrt{3}}{7}h $
38%
D
$ \frac{7}{\sqrt{3}}h $
4%
Solution:
$\frac{\sin 90^{\circ}}{\sin C}=\mu$
$\sin C=\frac{1}{\mu} \Rightarrow \frac{R}{\sqrt{R^{2}+h^{2}}}=\frac{3}{4}$
Squaring, $16 R^{2}=9 R^{2}+9 h^{2}$
$7 R^{2}=9 h^{2} \Rightarrow R=\frac{3}{\sqrt{7}} h$