Fig. shows an equiconvex lens (of refractive index `1.5`) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be `45.0 cm`. The liquid is removed and the experiment is repeated. The new distance is measured to be `30.0 cm`. What is the refractive index of the liquid ?

Explanation

Focal length of the convex lens, f1=30cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens+liquid), f=45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
`1/f= 1/f_(1)+1/f_(2)`
`1/f_(2)=1/f-1/f_(1)`
= `1/45-1/30=-1/90`
`therefore f_(2)=-90cm`
Let the refractive index of the lens be `mu_(1)` and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is -R
R can be obtained using the relation:
`1/(f_(1))=(mu_(1)-1)(1/R+1/-R)`
`1/30=(1.5-1)(2/R)`
`therefore R=(30)/(0.5 xx 2) = 30cm`

Let the refractive index of the liquid.
Radius of curvature of the liquid on the side of the lens, R=-30cm
The value of `mu_(2)` can be calculated using the relation.
`1/f_(2)=(mu_(2)-1)[1/-R-1/infty]`
`1/90 = (mu_(2)-1)[1/30-0]`
`mu_(2)-1 = 1/3`
`therefore mu_(2)=4/3=1.33`
Hence, the refractive index of the liquid is 1.33.