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Question

Physics Question on Ray optics and optical instruments

A point source of light is kept below the surface of water (nw=43)(n_w = \frac{4}{3}) at a depth of 7\sqrt{7} m. The radius of the circular bright patch of light noticed on the surface of water is

A

37m\frac{3} {\sqrt {7}} m

B

3m3\,m

C

73m\frac{\sqrt {7}} {3} \,m

D

7m{\sqrt {7}}\, m

Answer

3m3\,m

Explanation

Solution

When the ray of light is incident from water-air interface at critical angle (θc)\left(\theta_{c}\right), the refracted ray becomes parallel to the interface.

Hence, the radius of the circular bright patch of light noticed on the surface of water is given by
R=htanθcR=h \,\tan \,\theta_{c} (see figure)
=7sinθccosθc=7=\sqrt{7} \frac{\sin \theta_{c}}{\cos \theta_{c}}=\sqrt{7}
1/μ11μ2\frac{1 / \mu}{\sqrt{1-\frac{1}{\mu^{2}}}}
=7μ21=\frac{\sqrt{7}}{\sqrt{\mu^{2}-1}}
=3m=3\, m