**Problem 6.87: A cylindrical rod made of glass with refractive index $1.5$ is placed in air. It has a hemispherical end surface with radius $2.0\:cm$. A small object is placed at a distance of $6.0\:cm$ on the axis of the rod. Find the location of the image.**

**Solution:**

Here light falls from the object to the lens.

So that direction from left to right is taken as positive.

The object distance is measured from the lens to the object which is in the negative direction.

So object distance $u\:=\:-6.0\:cm$.

The center of curvature of the spherical surface is on the right side of the surface, so the radius is positive.

First medium is air. So $n_1\:=\:1$.

Second medium is glass. So $n_2\:=\:1.5$.

For refraction on a convex surface, we have $\frac{n_2}{v}\:-\:\frac{n_1}{u}\:=\:\frac{n_2\:-\:n_1}{R}$.

Substituting the given values,

$\frac{1.5}{v}\:-\:\frac{ 1}{-6.0}\:=\:\frac{1.5\:-\:1}{+2.0}$.

Or, $\frac{1.5}{v}\:=\:\frac{1}{4.0}\:-\:\frac{ 1}{6.0}\:=\:\frac{1}{12}$.

So image distance $v\:=\:18\:cm$.

The image is located in the direction of the light well within the glass rod.