A solid glass sphere of radius 10 cm and refractive index $\frac{3}{2}$ is placed in a medium of refractive index $\mu =2$ . The hemispherical portion of the sphere is silvered. A point object P is placed at distance 30 cm from the centre of sphere. The distance of final image of point object P from center of sphere is $\frac{50n}{11}$ cm. Find the value of n.

Image formation due to first refraction through unslivered surface.

 

$\begin{array}{l}{\mu }_1=2,{\mu }_2=\frac{3}{2}u=-20cm and r=+10cm\\ According to refractive surface formula, \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{r}\\ \Rightarrow \frac{3}{2v}-\frac{2}{-20}=\frac{\frac{3}{2}-2}{+10}=-\frac{1}{20}\\ \therefore v=-10cm\\ Thus image P_1 formed due to refraction through unsilvered surface \\ behaves as object for silvered surface. \\ The silvered surface behaves as concave mirror. \\ For concave mirror, \\ u\text{'}=-30cm\\ f\text{'}=\frac{-10}{2}=-5cm\\ According to mirror formula, \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\\ \Rightarrow \frac{1}{v}-\frac{1}{30}=-\frac{1}{5}\\ \therefore v=-6cm\\ The image P_2 formed due to reflection through silvered\\ surface behaves as object for second time refraction through unsilvered surface.\\ Here, {\mu }_1^{"}=\frac{3}{2},{\mu }_2^{"}=2, u"=-14cm,r"=-10cm\\ \frac{{\mu }_2^{"}}{v"}-\frac{{\mu }_1^{"}}{u"}=\frac{{\mu }_2^{"}-{\mu }_1^{"}}{r"}\\ \Rightarrow \frac{2}{v"}-\frac{3}{2(-14)}=\frac{2-\frac{3}{2}}{-10}=-\frac{1}{20}-\frac{3}{28}=\frac{-7-15}{140}\\ \therefore v"=-\frac{140}{22}\times 2=-\frac{140}{11}cm\\ Thus the position of final image P_3 from centre of sphere \\ is =10+\frac{140}{11}=\frac{110+140}{11}=\frac{250}{11}cm=\frac{50n}{11}cm\\ \because n=5\end{array}$