Question

In a medium of refractive index $1.6$ and having a convex surface, a point object is placed at a distance of $12\text{cm}$ from the pole. The radius of curvature is $6\text{cm}$. Locate the image as seen from air.

• A. A real image at $30\text{cm}$
• B. A virtual image at $30\text{cm}$
• C. A real image at $4.28\text{cm}$
• D. A virtual image at $4.28\text{cm}$

Answer 1

The correct option is B A virtual image at $30\text{cm}$


As per the sign convention (direction of incident ray taken as +ve)
Given that,
Radius of curvature, $R=-6\text{cm}$
Refractive index of the denser medium, ${\mu }_1=1.6$
Distance of object from pole, $u=-12\text{cm}$
At the spherical interface.

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\Rightarrow \frac{1.0}{v}-\frac{1.6}{-12}=\frac{1-1.6}{-6}$
or $\frac{1}{v}=\frac{1}{10}-\frac{2}{15}=\frac{3-4}{30}=\frac{-1}{30}$
$\therefore v=-30\text{cm}$
Image formed will be virtual at a distance $30\text{cm}$ on the same side as the object.

Why this question? Caution: Be cautions about the side at which the centre of curvature of a spherical surface is lying, to determine the sign of radius of curvature of spherical surfaces.