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.

Explanation

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`