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 surface `I` and `R_(1)` and refractive index of the lens be `mu`. Let parallel rays be incident on the lens.
Applying refraction formula at the first surface,
`(mu)/(V_(1))-(1)/(oo)=(mu-1)/(R_(1))` (i)
At the second surface `(2)/(V)=(mu)/(V_(1))=(2-mu)/(-20)` (ii)
Addinig (i) and (ii), we get
`(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=40 cm rArr f=40 cm`