An equiconcave diverging lens of focal length F is cut into two equal halves. The two halves are turned around and joined with some liquid between them.The lens obtained is converging with focal length F. If the refractive index of the liquid is 3 then what is the refractive index of the lens?

Options

A. 4
B. 2
C. 5
D. 1.5

Answer

b.

Explanation

For initial diverging lens,
`(1)/(-F)=(mu-1)((2)/(-R))`
`rArr (1)/(F)=((mu-1)2)/(R)` (i)
For final combination: `(1)/(F) =(1)/(f_(1))+(1)/(f_(2))+(1)/(f_(3))`
`rArr (2(mu-1))/(R)=(mu-1)[-(1)/(R)]`
`+(mu_(liq)-1)(2)/(R)+(mu-1)[-(1)/(R)]`
`rArr mu=(mu_(liq)+1)/(2)=(3+1)/(2)=2`