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

Let `f_(1)` be the focal length of convex lens, radius of curvature of each curved face is R.
`(1)/(f_(1))=(m-1){(1)/(R)-((1)/(-R))}=(mu-1)(2)/(R)`
`rArr f_(1)=(R)/(2(mu-1))=(R)/(2((3)/(2)-1))=R`
When the space between the lens and mirror is filled by water of refractive index `mu_(1)=4//3` then the focal length of liquid concave lens `f_(2)` is
`(1)/(f_(2))=(mu_(1)-1)(-(1)/(R)-oo)`
`rArr f_(2)=(-R)/(mu_(1)-1)=-(R)/(((4)/(3)-1))=-3R`
The combined focal length of lenses if `F_(1) =15 cm `
`:. (1)/(F)=(1)/(f_(1))+(1)/(f_(2))rArr (1)/(15)=(1)/(R)-(1)/(3R)=(3-1)/(3R)`
`3R=30rArr R=10cm`
In the second case,
` F_(2)=25cm, Let mu_(1)=mu_(2) `.
`:. (1)/(F_(2))=(1)/(f_(1))+(1)/(f_(2))rArr (1)/(25)=(1)/(10)+(1)/(f_(2)^('))`
`rArr (1)/(f_(2)^('))=(1)/(25)-(1)/(10)=(2-5)/(50)`
`:. f_(2)^(')=(-50)/(3) cm`
`f'_(2)^(')=(R)/(mu_(2)-1)rArr mu_(2)-1=-(R)/(f_(2)^('))=(-10)/((-50//3))=(3)/(5)=0.6 `
`rArr mu_(2)=1+0.6=1.6 `