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.

Explanation

`(1)/(f)=((3)/(2)-1)((1)/(R)-(1)/(-R))=(1)/(R)`
`(1)/(f')=((4)/(3)-1)((1)/(-R)-(1)/(infty))=(1)/(-3R)`
The combination focal length
`(1)/(F)=(1)/(f)+(1)/(f')=(1)/(R)-(1)/(3R)=(2)/(3R)`
`F=(3R)/(2)`
Since image and object coincide i.e., after refraction from lenses, rays are incident normally on plane mirror i.e., rays are parallel to principal axis and hence rays are coming from combined focus of lenses.
`F=(3R)/(2)=15impliesR=10cm`
`(1)/(f)=(1)/(R)=(1)/(10)`
`(f)/(f')=(mu_0-1)((1)/(-R)-(1)/(infty))=-(mu_0-1)/(10)`
The combined focal length
`(1)/(F')=(1)/(f)+(1)/(f')=(1)/(10)-(mu_0-1)/(10)=(2-mu_0)/(10)`
`(1)/(25)=(2-mu_0)/(10)implies2-mu_0=(10)/(25)=-.4`
`mu_(0)=2-0.4=1.6`