Question

A transparent sphere of refractive index ‘μ’ and radius of curvature ‘R’ is kept in air. A point object is placed at a distance ‘d’ from the surface of the sphere so that the real image is formed at the same distance ‘d’ from exactly opposite side of the sphere. The distance ‘d’ is ______.

विकल्प

• `mu/R`

• R(μ − 1)

• `R/((mu - 1))`

• `R/((mu + 1))`

Answer 1

A transparent sphere of refractive index ‘μ’ and radius of curvature ‘R’ is kept in air. A point object is placed at a distance ‘d’ from the surface of the sphere so that the real image is formed at the same distance ‘d’ from exactly opposite side of the sphere. The distance ‘d’ is `bbunderline(R/((mu - 1)))`.

Explanation:

Given: Radius of sphere = R

Refractive index of the medium = µ

We know,

`mu_2/v - mu_1/u = (mu_2 - mu_1)/r`

`mu/infty + 1/d = (mu - 1)/R`    ...(∵ v = ∞ and u = −d)

Solving for d, we get

d = `R/(mu - 1)`