The refracting angle of prism is ${\left(\frac{14}{11}\right)}^0$ and refractive index is 1.5. A thin convex lens of focal length 20 cm is placed at distance 30 cm from the point source S as shown in figure. The position of image of point source S is

$\begin{array}{l}\because \delta =(\mu -1)A\\ =(1.5-1)\frac{14}{11}\times \frac{\pi }{180}=\frac{1}{2}\times \frac{14}{11}\times \frac{22}{7\times 180}=\frac{1}{90}\\ \therefore \tan \delta =\delta =\frac{h}{15}\\ \Rightarrow \frac{1}{90}=\frac{h}{15}\\ \therefore h=\frac{1}{6}cm\end{array}$

S’ behaves as object for the thin convex lens. For the lens,$u=-30cm,f=+20cm$

$\begin{array}{l}\therefore \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \because \frac{1}{v}-\frac{1}{-30}=\frac{1}{20}\\ Or \frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{3-2}{60}\\ \therefore v=60cm\\ \therefore h^{\text{'}}=-2\times \frac{1}{6}=\frac{-1}{3}cm [\because \frac{h^{\text{'}}}{h}=\frac{v}{u}]\end{array}$