Consider a point object $O$ situated at a distance of $30 \ cm$ from the centre of a sphere of radius $2 \ cm$ and refractive index $\mu_2 = 1.5$ as shown in the figure. If the refractive index of the region surrounding this sphere is $\mu_1 = 1.4$, then the position of the image due to refraction by the sphere with respect to the centre is:

$A$ laser beam falls on a crystal ball of radius $R$ as shown in the figure. What is its refractive index?

In a thin spherical fish bowl of radius $10 \, cm$ filled with water of refractive index $\mu = \frac{4}{3}$,there is a small fish at a distance of $4 \, cm$ from the centre $C$ as shown in the figure. Where will the image of the fish appear if seen from $E$ (in $, cm$)?

$A$ spherical glass is attached to a rigid wall as shown in the figure. An observer located at point $O$ is looking at a point $A$ on the wall. The refractive index of the glass is $1.5$ and that of air is $1.0$. The distances are $OA = 8 \text{ cm}$, $XA = 3 \text{ cm}$. If the radius of curvature of the spherical glass surface is $R = 5 \text{ cm}$, then the apparent distance of $A$ from the observer $O$ is (in $\text{ cm}$)

$A$ parallel beam of light travelling in water (refractive index $\mu_1 = 4/3$) is refracted by a spherical air bubble of radius $R = 2 \, cm$ situated in water. Assuming the light rays to be paraxial,the position of the image due to refraction at the first surface is:

$A$ spherical vessel of radius $10 \, cm$ is filled with water of refractive index $\mu = 4/3$. $A$ fish is at a distance of $4 \, cm$ from the center as shown in the figure. If viewed from end $E$,at what distance will the fish appear to be (in $, cm$)? (Neglect the thickness of the vessel)