A luminescent object is placed at a depth of ‘d’ in a (optically) denser medium of refractive index ‘μ’. Prove that radius r of the base of the cone of light, from the object, that can emerge out from the surface is $r = \frac{d}{\sqrt{μ^{2} - 1}} .$

Solution

A luminescent object is placed at a depth of 'd' in an optically denser medium of refractive index '

μ

Cone of light is formed due to total internal reflection.

∴ For total internal reflection to take place,

\sin i_{c} = \frac{1}{μ}

and, from the fig.,

\tan i_{c} = \frac{r}{d}

∴

\frac{\sin i_{c}}{\sqrt{1 - \sin^{2} i_{c}}} = \frac{r}{d}

\Rightarrow

r = \frac{d}{\sqrt{μ^{2} - 1}} .

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