A sphere of refractive index $\mu =2$ has inner radius 10 cm and outer radius 20 cm. It is placed in air as shown in the figure. The distance through which point mark P appear to shift is ……. cm. 

For refraction through the first surface  ${\mu }_1=2,{\mu }_2=1,u=-10cm,r=+10cm$

According to refractive surface formula,

$\begin{array}{l}\because \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{r}\\ \Rightarrow \frac{1}{v}-\frac{2}{-10}=\frac{1-2}{10}\\ \Rightarrow \frac{1}{v}=-\frac{1}{10}-\frac{2}{10}=\frac{-1-2}{10}=-\frac{3}{10}\\ \because v=-\frac{10}{3}cm\end{array}$

The image $P_1$ formed by refraction through first surface behaves as object for second surface.

            For second surface,  ${\mu }_1^{\text{'}}=1,{\mu }_2^{\text{'}}=2$

$\begin{array}{l}u\text{'}=-(20+\frac{10}{3})=-\left(\frac{70}{3}\right)cm=\frac{-70}{3}cm,\\ r\text{'}=-10cm\\ \because \frac{{\mu }_2^{\text{'}}}{v\text{'}}-\frac{{\mu }_1^{\text{'}}}{u\text{'}}=\frac{{\mu }_2^{\text{'}}-{\mu }_1^{\text{'}}}{r\text{'}}\\ \Rightarrow \frac{2}{v\text{'}}-\frac{1}{-\frac{70}{3}}=\frac{2-1}{-10}\\ \therefore v\text{'}=-14cm\end{array}$

For refraction through third surface ,

  $\begin{array}{l}{\mu }_1^{\text{'}}=2,{\mu }_2^{"}=1,{\mu }^{"}=-(14+10)=-24cm\\ \because r"=-20cm\\ \because \frac{{\mu }_2^{"}}{v"}-\frac{{\mu }_1^{"}}{u"}=\frac{{\mu }_2^{"}-{\mu }_1^{"}}{r"}\\ \Rightarrow \frac{1}{v"}-\frac{2}{-24}=\frac{1-2}{-20}\\ \Rightarrow v"=-30 cm\\ Thus, shifting in object = 40-30 = 10 cm\end{array}$