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?

Options

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

b

Explanation

(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`