A point source of light is placed 4 , m below | vedclass

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(D) The light from the point source will be cut off if the disc covers the area corresponding to the critical angle $	heta_c$ of the water-air interf

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The light from the point source will be cut off if the disc covers the area corresponding to the critical angle $	heta_c$ of the water-air interface.For total internal reflection to occur,the radius $r$ of the disc is given by the formula $r = \frac{h}{\sqrt{\mu^2 - 1}}$,where $h = 4 \, m$ is the depth and $\mu = 5/3$ is the refractive index.Substituting the values: $r = \frac{4}{\sqrt{(5/3)^2 - 1}} = \frac{4}{\sqrt{25/9 - 1}} = \frac{4}{\sqrt{16/9}} = \frac{4}{4/3} = 3 \, m$.The diameter of the disc is $D = 2r = 2 	imes 3 \, m = 6 \, m$.

$A$ point source of light is placed $4 \, m$ below the surface of water of refractive index $5/3$. The minimum diameter of a disc that should be placed over the source on the surface of water to cut off all light coming out of the water is ...... $m$.

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