What curvature must be given to the bounding surface of `mu = 1.5` for virtual image of an object in the medium of `mu = 1 at 10 cm` to be formed at a distance of `40 cm`. Calculate power of the refracting surface and also two principal focal lengths of the surface.

Explanation

`R = ?, u_2 = 1.5, mu_1 = 1`
`u = -10 cm, v = -40 cm`
As object is placed in rarer medium,
`:. -(mu_1)/(u)+(mu_2)/(v)=(mu_2 - mu_1)/( R)`
`-(1)/(-10)+(1.5)/(-40)=(1.5 - 1)/( R) =(1)/(2 R)`
`(1)/(2 R)=(1)/(10)-(3)/(80)=(5)/(80)=(1)/(16)`
`R = (16)/(2) = 8 cm`
As `R` is positive, the refracting surface is convex. Power of surface,
`P = (mu_2 - mu_1)/( R) = (1.5 -1)/(8//100) = 6.25 D`
For power in dioptre, `R` must be in metre.
Again, `f_1 = -(mu_1 R)/(mu_2 - mu_1) = -(1 xx 8)/(1.5 - 1) = -16 cm`,
and `f_2 = (mu_2 R)/(mu_2 - mu_1) = (1.5 xx 8)/(1.5 - 1) = 24 cm`.