One end of a cylindrical glass rod shown in figure, is ground to a hemispherical surface of radius `R=20mm.`

a. Find the image distance of a point object on the axis of the rod, 80mm to the left of the vertex. The rod is in air.
b. Let the same rod be immersed in water of refractive index `4//3` , the other quantities having the same values as before. Find the image distance.

Explanation

a. `mu_(1)=1, mu_(2)=1.5, R=+2mm, u=+80mm`
`(1.5)/(v)-(1)/(8)=(1.5-1)/(-20)`
`v^(')=+120mm`
The image is therefore formed at the right of the vertex.
( v is positive and at a distance of 120 mm from it.)
b. `(1.5)/(v)+(-1.33)/(-80)=(1.5-1.33)/(+20), v=-180mm`
The fact theat v is negatvie means that the rays after refraction by the surface are not converging but appear to diverge from a pint 180mm to the left of the vertex. In the illustration, then, the surface forms a virtual image 180mm to the left of the vertex.