Question

A parallel narrow beam of light is incident on the surface of a transparent hemisphere of radius $R$ and refractive index $\mu =1.5$ as shown. The position of the image formed by refraction at the spherical surface only is

• A. $\frac{R}{2}$
• B. $3R$
• C. $\frac{R}{3}$
• D. $2R$

Answer 1

The correct option is B $3R$


Sign convention: Direction of incident ray taken as +ve.
For the spherical interface, the object is at infinity.
$u=-\infty$ ($\because$ incident rays are parallel & horizontal to the principle axis)
Radius of curvature, $R=+R$
By using formula
$\Rightarrow \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
The incident rays are coming from air medium i.e. ${\mu }_1=1,{\mu }_2=1.5$
or, $\frac{1.5}{v}-\frac{1}{-\infty }=\frac{1.5-1}{+R}$
or, $\frac{3}{2v}+0=\frac{1}{2R}$
$\therefore v=3R$

Why this question? Caution: In the question, it has been asked "to find position of image formed by reflection at spherical surface". Thus angle $\theta ={45}^{\circ }$ for plane surface is meaningless here.