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(C) The light from the source $S$ at the bottom will emerge from the water surface only if the angle of incidence is less than or equal to the critica

Vedclass · Accepted Answer

$\frac{6}{\sqrt{7}} \ m$

Vedclass · Answer

(C) The light from the source $S$ at the bottom will emerge from the water surface only if the angle of incidence is less than or equal to the critical angle $C$. The radius $r$ of the circular disc formed on the surface is given by $r = h 	an C$. Since $\sin C = \frac{1}{\mu}$,we have $	an C = \frac{1}{\sqrt{\mu^2 - 1}}$. Substituting the given values $h = 1 \ m$ and $\mu = \frac{4}{3}$: $r = 1 	imes \frac{1}{\sqrt{(\frac{4}{3})^2 - 1}} = \frac{1}{\sqrt{\frac{16}{9} - 1}} = \frac{1}{\sqrt{\frac{7}{9}}} = \frac{3}{\sqrt{7}} \ m$. The diameter of the disc is $D = 2r = 2 	imes \frac{3}{\sqrt{7}} = \frac{6}{\sqrt{7}} \ m$.

$A$ wide container is filled with water $\left(\mu=\frac{4}{3}\right)$ up to a height of $1 \ m$. Find the diameter of the disc at the top of the water surface from which light is coming out.

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