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(C) For the source $S$ to be invisible from above,the light rays from $S$ must undergo total internal reflection at the liquid-air interface or be blo

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(C) For the source $S$ to be invisible from above,the light rays from $S$ must undergo total internal reflection at the liquid-air interface or be blocked by the opaque disc $D$. The condition for the source to be completely hidden is that the radius of the disc $r$ must be at least equal to the radius of the circle of illumination formed by the critical angle $	heta_c$ at the surface.The radius $r$ of the circle of illumination is given by:$r = h 	an 	heta_c$Since $\sin 	heta_c = 1/\mu$,we have $	an 	heta_c = \frac{1}{\sqrt{\mu^2 - 1}}$.Thus,$r = \frac{h}{\sqrt{\mu^2 - 1}}$.Given $r = 1\, cm$ and $\mu = 5/3$,we substitute these values:$1 = \frac{h}{\sqrt{(5/3)^2 - 1}}$$1 = \frac{h}{\sqrt{25/9 - 1}}$$1 = \frac{h}{\sqrt{16/9}}$$1 = \frac{h}{4/3}$$h = 4/3 = 1.33\, cm$.Therefore,the maximum height of the liquid is $1.33\, cm$.

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