Question

A glass sphere of radius 2R and refractive index n has a spherical cavity of radius R, concentric with it.

Q. When viewer is on right side of the hollow sphere, what will be apparent change in position of the object?
A. `((n-1))/((3n+1))R`,towards left
B. `((n+1))/((3n-1))R`,towards left
C. `((n+1))/((3n+1))R`,towards right
D. `((n-1))/((3n+1))R`,towards right

Answer 1

Correct Answer - b
(i) Viewer on the left of hollow sphere `:` Single refraction takes place at surface S. From the single surface refraction equation, we have
`(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`
(ii) When the viewer is on the right, two refractions take place at surface `S_(1)` and `S_(2)`.
For refraction at surface `S_(1) :`
`(n)/(v_(1))-(1)/((-2R))=((n-1))/((-R))`
which ono solving for `v_(1)` yields
`v_(1)=-(2nR)/(2n-1)`
The first lies to the left of `S_(1)` act 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`
`(1)/(v_(2))=-(n)/([(4n-1)/(2n-1)])=(1-n)/(-2R)`
On solving for `v_(2)` , we get
`v_(2)=-(2(4n-1))/(3(n-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`