Question

A beaker is filled with water as shown in Fig. (a). The bottom surface of the beaker is a concave mirror of large radius of curvature and small aperture. The height of water is $h=40\, cm$. It is found that when an object is placed $4\, cm$ above the water surface, the image coincides with the object. Now the water level $h$ is reduced to zero but there will still be some water left in the concave part of the mirror as shown in Fig. (b). Find the new height (in cm) of the object $h^{\prime}$ above the new water surface so that the image again coincides with the object. (Refractive index of water $=4 / 3$ )

Answer 1

The image coincides with object $O$ if the ray starting from $O$ is incident normally on mirror as shown in Fig. (a).
Here the image is located at $x=4 \times \frac{4}{3}=\frac{16}{3}$
In second case the image will coincides with object $O$ if ray starting from $O$ is incident normally on mirror as shown in Fig. (b).

Thus, here we use $\frac{136}{3}=h^{\prime} \times \frac{4}{3} $
$\Rightarrow h^{\prime}=\frac{136}{3} \times \frac{3}{4}=34 \,cm$