A thin equiconvex lens of refractive index `3//2` is placed on a horizontal plane mirror as shown in figure. The space between the lens and the mirror is filled with a liquid of refractive index `4//3` . It is found that when a point object is placed 15 cm above the lens on its priincipal axis, the object coincides with its own image.

Q. If another liquid is filled instead of water, the object and the image coincide at a distance 25 cm from the lens.
Calculate the refractive index of the liquid.

Answer

`1.6`

Explanation

In first case
`P_(eq)=2P_(1)+2P_(2)+0`
`P_(eq)=2(P_(1)+P_(2))`
`P_(eq)=2[(2(underset(a)""mu_(g)-1))/(R )-((underset(a)""mu_(l)-1))/(R )]" " implies P_(eq) = 2[(2((3)/(2)-1))/(R)- (((4)/(3)-1))/(R)]`
`P_(eq)=2[(1)/(R )-(1)/(3R)] rArrP_(eq)(4)/(3R)`
`(1)/(F_(eq))=(4)/(3R)....(1)`
since the image is formed at object itself is means object is placed at`R=15cm`
`F_(eq)=(-15)/(2)...(2)`
from(1)&(2)
`(2)/(15) =(4)/(3R )rArr R=10cm`
For second case
`P'_(eq)=2P_(1)+2P_(2)'+0`
`(1)/(-F_(eq)^('))=d[(1)/(R )-((underset(a)""mu_(l)-1))/R] rArr-(2)/(-25)=2[(2-underset(a)""mu_(l))/(R )]`
`2-underset(a)""mu_(l)=(R )/(25)rArr underset(a)""mu_(l) =2-(R )/(25)`
`=2-(10)/(25)`
`underset(a)""mu_(l)=1.6`