Question

A thin equiconvex lens of refractive index $3/2$ is placed on a horizontal plane mirror as shown in figure. The space between the lens and the mirror is filled with a liquid of refractive index $4/3$. It is found what when a point object is placed $15cm$ above the lens on its principal axis, the object coincides with its own image.
The radius of curvature of the convex surface is

• A. $10cm$
• B. $15cm$
• C. $20cm$
• D. $25cm$

Answer 1

Answer: (A)

$10cm$

The setup would be detailed as shown above. A thin layer of air would exist between mirror and liquid. Water could be treated as a plano-concave lens of refractive index $4/3$.

From the principle of reversibility of light we can say that if the final image is to be formed on the object itself, the image after refraction from liquid to air would be formed on the mirror itself. If the mirror and that image have a finite distance, the ray cannot be traced back to the same point after reflection from the mirror.

Consider refraction from first surface,

$\frac{3}{2v_1}-\frac{1}{-15}=\frac{\frac{3}{2}-1}{R}$

Refraction from second surface,

$\frac{4}{3v_2}-\frac{3}{2v_1}=\frac{\frac{4}{3}-\frac{3}{2}}{-R}$

Refraction from liquid to air,

$\frac{1}{v_3}-\frac{4}{3v_2}=\frac{1-\frac{4}{3}}{\infty }$

Hence, $\frac{1}{v_3}=\frac{1}{-15}+\frac{2}{3R}$

Since $v_3=0$

Hence $R=10cm$