Question

A glass hemisphere of refractive index $4/3$ and radius $4\text{cm}$ is placed on a plane mirror. A point object is placed at distance ${}^{\prime }{d}^{\prime }$ on the axis of this sphere as shown. If the final image is at infinity, then the value of ${}^{\prime }{d}^{\prime }$ in $\text{cm}$ is

Answer 1

The image of object $O$ due to refraction at the plane surface is formed at $I$ such that
$\Rightarrow AI=\frac{4}{3}d$


$I$ acts as the object for the curved surface. The curved surface makes image of $I$ at $I\text{'}$.
Using curved surface refraction formula,
$\frac{1}{v}-\frac{\frac{4}{3}}{-(R+\frac{4}{3}d)}=\frac{1-\frac{4}{3}}{-R}$
or $\frac{1}{v}=\frac{1}{3R}-\frac{4}{3R+4d}$

$I\text{'}$ acts as the object for the plane mirror. Mirror makes its image at $I\text{'}\text{'}$ distant $v$ above $B$.

$I\text{'}\text{'}$ acts as virtual object for the curved surface which makes its image at infinity.
Then, $\frac{4/3}{\infty }-\frac{1}{v}=\frac{4/3-1}{R}$
$\Rightarrow \frac{\frac{4}{3}}{\infty }-[\frac{1}{3R}-\frac{4}{3R+4d}]=\frac{\frac{1}{3}}{R}$

On solving, we get
$d=\frac{3}{4}R=\frac{3}{4}\times 4=3\text{cm}$

There is no further refraction at plane surface since the image is at infinity (parallel rays normal to the plane surface)