Question

A glass shere of radius 2R, refractive index n has a spherical cavity of radius R, concentric with it. A black spot on the inner surface of the hollow sphere is viewed from the left as well as right. Obtain the shift in position of the object.

Answer 1

a. Viewer on the left of hollow sphere: Single refraction takes place at surface S.
From the single surface refraction equation, we have
`mu_(2)=1, mu_(1)=n, u=-R, and R=-2R`
`(1)/(v)-(n)/((-R))=((1-n))/((-2R))`
which on solving for v yields, `v=-((2R)/(n+1))`
Image is on the right of refracting surface S.
Shift= Real depth-Apparent depth
`R-((2R)/(n+1))=((n-1))/((n+1))R`
b. When the viewer is on the right, two rerfractions take place at surfaces `S_(1)` and `S_(2)`.
`mu_(2)=n, mu_(1)=1, u=-2R, and R=-R`
For refraction at surface `S_(1): (n)/(v_(1))-(1)/((-2R))=((n-1))/((-R))`
which on solving for `v_(1)=-(2nR)/(2n-1)` .
The first lies to the left of `S_(1) ` and acts as object for refraction at the second surface. We have to shift the origin of cartesian coordinate system to the vertex of `S_(2)`. The object distance for the second surface is
`u_(2)=-[(2nr)/(2n-1)+R]=-((4n-1)/(2n-1))R`
Here `mu_(2)=1, mu_(1)=n, R=-2R`
`rArr (1)/(v_(2))-(n)/(-[(4n-1)/(2n-1)]R)=(1-n)/(-2R)`
On solving for `v_(2)` , we get `v_(2)`=`-(2(4n-1))/((3n-1))R`
The minus sign shows that image is virtual and lies to the left of `S_(2)` .
Shift= Real depth-Apparent depth
`=3R-(2(4n-1)R)/((3n-1))`
`=((n-1))/((3n-1))R`