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(A) Let the refractive index of the glass be $\mu.$ When the light beam is at a distance $d$ from the central line,it strikes the curved surface at an

Vedclass · Accepted Answer

$\frac{R}{d}$

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(A) Let the refractive index of the glass be $\mu.$ When the light beam is at a distance $d$ from the central line,it strikes the curved surface at an angle of incidence $	heta.$ From the geometry of the triangle formed by the radius $R$ and the distance $d,$ we have $\sin 	heta = \frac{d}{R}.$ However,the angle of incidence at the curved surface is the angle between the incident ray and the normal. The normal at the point of incidence passes through the centre of the semi-circle. Looking at the geometry,the angle $	heta$ between the normal and the vertical incident ray is such that $\sin 	heta = \frac{d}{R}.$ For the light to not exit the cylinder,it must undergo total internal reflection at the curved surface. This happens when the angle of incidence $	heta$ is equal to the critical angle $C.$ Thus,$\sin C = \frac{1}{\mu}.$ Since $\sin 	heta = \frac{d}{R}$ and $	heta = C,$ we have $\frac{d}{R} = \frac{1}{\mu}.$ Therefore,the refractive index is $\mu = \frac{R}{d}.$

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