For an object placed at infinity, the image after first refraction will be formed at `v_(1)`
`:. (mu_(2))/(v_(1))-(mu_(1))/(-oo)=(mu_(2)-mu_(2))/(+R)` (i)
The image second refraction will be found at `v_(2)`
`:. (mu_(3))/(v_(2))-(mu_(2))/(v_(1))=(mu_(3)-mu_(1))/(+R)` (ii)
Adding (i) and (ii) `(mu_(3))/(v_(2))=(mu_(3)-mu_(1))/(R)`
`rArr v_(2)=(mu_(3)R)/(mu_(3)-mu_(1))`
Therefore, focal length will be `(mu_(3) R)/(mu_(3)-mu_(1))` .