An object is placed `50 cm` from the surface of a glass sphere of radius `10 cm` along the diameter. Where will the final image be formed after refraction at both the surfaces ? `mu` of glass `=1.5`.

Answer

`20 cm`

Explanation

Here, `u = - 50 cm, R = 10 cm, mu_(1) = 1, mu_(2) = 1.5`
Refraction at surface `P_(1)A`
Virtual image at `I_(1)` where `P_(1)I_(1) = v_(1)`
`:. -(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`- (1)/(- 50) + (1)/(v_(1)) = (1.5 - 1)/(10) = (1)/(20)`

`(3)/(2 v_(1)) = (1)/(20) - (1)/(50) = (3)/(100), v_(1) = 50 cm`
Refraction at surface `P_(2)B`
`I_(1)` acts as virtual object
`u = P_(2)I_(1) = P_(1)I_(1) - P_(1)P_(2) = 50 - 20 = 30 cm`,
`v_(2) = P_(2)I = ?, R = - 10 cm`
`:. -(mu_(2))/(u) + (mu_(1))/(v) = (mu_(1) - mu_(2))/(R )`
`-(1.5)/(30) + (1)/(v) = (1 - 1.5)/(-10) = (1)/(20)`
`(1)/(v) = (1)/(20) + (3)/(60) = (1)/(60), v = 10 cm`.
Distance of final image `I` from centre of sphere
`CI = CP_(2) + P_(2)I = 10 + 10 = 20 cm`