Question

Consider a transparent hemisphere $(n=2)$ in front of which a small object is placed in air $(n=1)$ as shown.
What is the nature of final image of the object when $x=2R$?

• A. Erect and magnified
• B. Inverted and magnified
• C. Erect and same size
• D. Inverted and same size

Answer 1

The correct option is D Inverted and same size

For refraction from first surface,

$\frac{2}{v_1}-\frac{1}{-2R}=\frac{1}{R}$

Hence $v_1=4R$

Magnification from first refraction= $-\frac{{\mu }_i}{{\mu }_r}\frac{v_1}{u_1}=-\frac{1}{2}\frac{4R}{-2R}=1$

For refraction from second surface,

$\frac{1}{v_2}-\frac{2}{3R}=\frac{-1}{\left(\infty \right)}$

Hence, $v_2=\frac{3R}{2}$

Magnification from second surface= $-\frac{2}{1}\frac{\frac{3R}{2}}{3R}=-1$

Net magnification = Magnification from 1$\times$Magnification from 2

= $-1$

Hence final image is inverted and same size.