Two similar thin equi-convex lenses, of focal f each, are kept coaxially in contact with each other such that the focal length of the combination is `F_(1)`, When the space between the two lens is filled with glycerin (which has the same refractive index `(mu=1.5)` as that of glass) then the equivlent focal length is `F_(2)`, The ratio `F_(1): F_(2)` will be

Options

A. `3:4`
B. `2:1`
C. `1:2`
D. `2:3`

Answer

C

Explanation

According to lens maker.s formula
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(f)=(mu-1)((1)/(R)-(1)/(-R))=(1.5-1)((2)/(R))=(1)/(R)`
Two similar equi-convex lenses of focal length f each are held in contact with each other. The focal length `F_(1)` of the combination is given by
`(1)/(F_(1))=(1)/(f)+(1)/(f)=(2)/(f)`

`F_(1)=(f)/(2)=(R)/(2)` ...(i)
For glycerin in between lenses, there are three lenses, one concave and two convex.
Focal length of the concave lens is given by
`(1)/(f.)=(1.5-1)((-2)/(R))=-(1)/(R)`
Now, equivalent focal length of the combination is,
`(1)/(F_(2))=(1)/(f)+(1)/(f.)+(1)/(f)`
`(1)/(F_(2))=(1)/(R)-(1)/(R.)+(1)/(R)=(1)/(R)`
`F_(2)=R`
Dividing equation (i) by (ii), we get
`(F_(1))/(F_(2))=(1)/(2)`