A solid hemisphere of refractive index mu_1 = | vedclass

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(A) Let the refractive index of the hemisphere be $\mu_1 = 3/2$ and the special glass be $\mu_2$. The object $P$ is at a distance $R$ from the flat su

Vedclass · Accepted Answer

The refractive index $\mu_2$ is $2$.

Vedclass · Answer

(A) Let the refractive index of the hemisphere be $\mu_1 = 3/2$ and the special glass be $\mu_2$. The object $P$ is at a distance $R$ from the flat surface $C$. $1$. Refraction at the flat surface of the hemisphere: The object $P$ is at distance $R$ from the surface. The image $I_1$ formed by refraction at the flat surface is at distance $v_1$ from the surface. Using $\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_{surface}}$,where $R_{surface} = \infty$,we get $\frac{\mu_2}{v_1} = \frac{\mu_1}{R} \Rightarrow v_1 = \frac{\mu_2 R}{\mu_1}$. $2$. Reflection at the silvered surface $MN$: The distance of $I_1$ from the silvered surface is $d = t + v_1 = \frac{2R}{3} + \frac{\mu_2 R}{\mu_1}$. The mirror forms an image $I_2$ at distance $d$ behind the mirror. $3$. Refraction back through the flat surface: The light travels back through the glass of thickness $t$ and then through the hemisphere. For the final image to be at $C$,the rays must emerge from the flat surface as if coming from $C$. Using the condition for the final image to be at $C$,the effective distance of the object from the flat surface after reflection must be such that the refraction at the flat surface results in an image at $C$. Calculations show that for the image to be at $C$,$\mu_2 = 2$.

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