In the figure shown,$O$ is the centre of the glass sphere. When the spot $P$ on the sphere is viewed almost normally,it appears:

$A$ small object is placed in the air, at a distance of $45 \,cm$ from a convex refracting surface of radius of curvature $15 \,cm$. If the surface separates air from glass of refractive index $1.5$, then the position of the image is: (in $\,cm$)

Light from a point source in air falls on a convex spherical glass surface $(\mu = 1.5, R = 20 \, cm)$. The distance of the light source from the glass surface is $100 \, cm$. At what position is the image formed (in $, cm$)?

Water (with refractive index $\mu = \frac{4}{3}$) in a tank is $18 \ cm$ deep. Oil of refractive index $\mu = \frac{7}{4}$ lies on the water,forming a convex surface of radius of curvature $R = 6 \ cm$ as shown. Consider the oil to act as a thin lens. An object $S$ is placed $24 \ cm$ above the water surface. The location of its image is at $x \ cm$ above the bottom of the tank. Then $x$ is

Derive the relation between object distance $(u)$,image distance $(v)$,refractive indices of the media ($n_1$ and $n_2$),and the radius of curvature $(R)$ for a spherical refracting surface.

$A$ beam of diameter $d$ is incident on a glass hemisphere as shown. If the radius of curvature of the hemisphere is very large in comparison to $d$,then the diameter of the beam at the base of the hemisphere will be: