A thin bi-convex lens made up of glass of refractive index `3//2` is placed is front of a plane mirror. The space between the lens and the mirror is filled with water of refractive index `4//3` . The radii of curvatures of the lens are `R_(1)=15cm` and `R_(2)=25 cm ` . A point object is placed at distance x from the surface whose curvature is `R_(1)`. The distance x is greater than the separation between the lens and the mirror. Find the value of x so that image must coincide with the object.

Explanation

Refraction from first surface `u=xrArr v=?rArr R=15cm`
`mu_(1)=1(air)rArr mu_(2)=(3)/(2)`
`:. (mu_(2))/(v)-(mu_(1))/u=(mu_(2)-mu_(1))/(R)rArr (3)/(2V)+(1)/(x)=(1)/(2xx15)`
`rArr (3)/(2V)=-(1)/(x)+(1)/(30)` (i)
Refraction from second surface
`u=VrArr V^(')=? rArr R=-25cm`
`mu_(1)=(3)/(2)rArr mu_(2)=(4)/(3)`
To coincide the final image to the object, the reflected ray must retrace the path
`rArr V^(') =oo`
`rArr (4)/(3V^('))-(3)/(2V)=((4)/(3)-(3)/(2))/(-25)rArr (3)/(2V)=(1)/(150)` (ii))
From (i) and (ii)
`rArr-(1)/(x)+(1)/(30)=(1)/(150)rArr x=25cm`