Question

A plane mirror is suspended vertically at the center of a large thin-walled spherical flask filled with water. The diameter of the flask is 10 inches. An observer whose eye is 35 inches from the mirror as shown in figure tries to see an image of his own eye. What is distance of the image from the eye (in inches). The effect of the thin glass walls of the flask may be neglected. Take ${\mu }_{\text{water}}=\frac{4}{3}$.

Answer 1

Using formula for refraction at curved surface
$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$
For first refraction
$n_1=1,n_2=\frac{4}{3},u=-30in$
$\frac{4}{3v}-\frac{1}{-30}=\frac{\frac{4}{3}-1}{5}$
$\frac{4}{3v}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}$
$v={40}^{\prime \prime }$

Image due to first refraction is formed at 40 inch from curved surface. So, object distance for reflection at plane mirror is 35 inch towards right. Hence image distance is 35 inch towards left from plane mirror.

For final refraction at curved surface:
As image due to refraction is 35 inch towards left from plane mirror, object distance for final refraction at curved surface can be considered as 30 inch.
hence,
$\frac{1}{v}-\frac{4/3}{+30}=\frac{1-4/3}{-5}$
$v=9^{\prime \prime }$
Final image is formed at 9 inch from curved surface towards left.
Distance from observer $=30-9={21}^{\prime \prime }$