In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is R. Determine the focal length of this system.

Explanation

(i) For refraction at the first surface of the lens
Here, `u = - oo, R = + R`
Suppose the first surface converges the incident parallel beam at a distance `v_(1)`, assuming medium `mu_(2)` to be continuous.
As `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`:. -(mu_(1))/(-oo) + (mu_(2))/(v_(1)) = (mu_(2)-mu_(1))/(R )`
or `(mu_(2))/(v_(1)) = (mu_(2) - mu_(1))/(R )` ...(i)
(ii) For refraction at second surface of the lens
`-(mu_(2))/(u) + (mu_(3))/(v) = (mu_(3) - mu_(2))/(R )` ...(ii)
The image formed by the first surface acts as virtual object for the second surafce and this second surface forms the image at the focus of the lens.
`:. u = v_(1), v = + f and R = + R`
From (ii), `-(mu_(2))/(v_(1)) + (mu_(3))/(+f) = (mu_(3) - mu_(2))/(R )` ...(iii)
Adding (i) and (ii), we get
`(mu_(3))/(f) = (mu_(2) - mu_(1))/(R ) + (mu_(3) - mu_(2))/(R ) = (mu_(3) - mu_(1))/(R )`
`f = (mu_(3)R)/(mu_(3) - mu_(1))`