Focal length of a thin lens in air is 10cm. Now, medium on one side of the lens is replaced by a medium of refractive inidex `mu=2`. The radius of curvature of the surface of lens, in contact with the medium, is 20cm. Find the new focal length.

Explanation

Let radius of `1` surface be `R_(1)` and refreactive index of lens be `mu`. Let parallel rays be incident on the lens. Appyling refractiion formula at first surface
`(mu)/(v_(1)) - (1)/(oo = (mu-1)/(R_(1)) .....(1)`
At `II` surface `(2)/(v) - (mu)/(v_(1)) = (2-mu)/(-20) ....(2)`
Adding (1) and (2)
`(mu)/(v_(1)) - (1)/(oo) + (2)/(v) - (mu)/(v_(1)) = (mu-1)/(R(1)) + (2-mu)/(-20)`
`= (mu-1)((1)/(R_(1)) - (1)/(-20)) = (mu-1)/(20) - (2-mu)/(20) = (1)/(f)` (in air) `+ (1)/(20) - (2)/(20)`
`rArr v = 40cm rArr f = 40cm`