An object is placed `10cm` away form a glass piece `(n = 1.5)` of length `20cm` bounded by spherical surfaces of radii of curvaure `10cm`. Find the position of final image formed after two refractions at the spherical surfaces.

Answer

`50cm` right of `B`

Explanation

For first reaction:
`(1.5)/(v_(1)) - (1)/(-10) = (1.5-1)/(10) rARr v_(1) = -30cm`
For secound refractions
`u = -(30+20) = -50cm`
`:. (1)/(v_(2)) - (1.5)/(-50) = (1-1.5)/(-10)`
`rArr v_(2) = 50cm`
Hence, final image is formed `50cm` right of `B`.