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

Here, refractive index of convex lens, `mu = 1.5`.
Let `mu_(1)` be the refractive index of liquid.
When the liquid is present, the system behaves as a combination of the convex lens `(mu)` and plano-concave lens of liquid `(mu_(1))`.
So, in this case, the distance of the needle from the lens, for which the inverted image is found at the position of needle itself, is equal to the combined focal length of the convex lens and plano-concave lens of liquid.
`therefore" "f = 45 cm`
When liquid is removed, then `f_(1)=30cm` is focal length of convex lens only. If `f_(2)` is focal length of plano - concave liquid lens, then from `(1)/(f_(1))+(1)/(f_(2))=(1)/(f)`
`(1)/(f_(2))=(1)/(f)-(1)/(f_(1))`
`=(1)/(45)-(1)/(30)=(2-3)/(90)`
`"or "(1)/(f_(2))=(-1)/(90)`
`f_(2)=-90cm`
Using lens maker.s formula for equiconvex lens where in
`R_(1)=R and R_(2)=-R`
`(1)/(f_(1))=(mu-1)((1)/(R_(1))+(1)/(R_(2)))`
`(1)/(30)=(1.5-1)((1)/(R)+(1)/(R))`
`=(1)/(2)xx(2)/(R )=(1)/(R )`
`R=30cm`
For plano - concave lens of liquid.
`f_(2)=-90cm, mu=?`
`R_(1)=-30cm" (for plane surface)"`
`R_(2)=oo`
`"As "(1)/(f_(2))=(mu_(1)-1)((1)/(R_(1))-(1)/(R_(2)))`
`therefore" "-(1)/(90)=(mu_(1)-1)((1)/(-30)-(1)/(oo))`
`-(1)/(90)=(mu_(1)-1)/(-30)`
`mu_(1)-1=(-30)/(-90)=(1)/(3)`
`mu_(1)=(1)/(3)+1=(4)/(3)=1.33`