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A glass sphere of radius 2R and refractive index n has a spherical cavity of radius R, concentric with it. Q. When viewer is ono left side of the hollow sphere, what will be the shift in position of the object? |
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Answer» `((n+1))/((n-1))R`,right `(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` |
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