A convex lens has its radi of curvature equal. The focal length of the lens is `f`. It is is dividded vertically into two identical plano-convex lenses by cutting it, then the focal length of the plane -convex lens is (`mu=` the refractive index of the material of the lens)

Options

A. `f`
B. `f/2`
C. `2f`
D. `(mu-1)f`

Answer

C

Explanation

given `R_(1)=R,R_(2)=-R`

when convex lens is complete `f=F`
Lens Maker's formula
`1/F=(mu-1)[1/(R_(1))-1/(R_(2))]`
`1/f=(mu-1)[1/R+1/R]`
`implies f=(R/(2mu-1))`
`R=2f(mu-1)`……i
Now, it is divided vertically into two identical plano convex lens

`1/(f')=(mu-1)[1/(R_(1))-1/(R_(2))]`
`[ :' R_(1)=R,R_(2)=oo]`
`1/(f_(1))=(mu-1)[1/R=1/oo)]`
`impliesf_(1)=R/((mu-1))=(2f(mu-1))/((mu-1))`
`f_(1)=2f`