The convex surface of a thin concave-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image is formed at the same place ? (b) If the concave part is filled with water (mu = 4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

Explanation

`(1)/(f_l)=((3)/(2)-1)((1)/(-60)-(1)/(-20))=((1)/(2))((-1+3)/(60))=(1)/(60)`
The image by lens should be formed as centre of curvature of mirror:
`(1)/(v)-(1)/(u)=(1)/(f_l)`
`(1)/(-20)-(1)/(-x)=(1)/(60)implies(1)/(x)=(1)/(60)+(1)/(20)=(1)/(15)impliesx=15cm`
Alternatively:
`(1)/(F)=(1)/(f_l)+(1)/(f_m)+(1)/(f_l)=(2)/(f_l)+(1)/(f_m)`
`=(2)/(60)+(1)/((20)/(2))=(1)/(30)+(1)/(10)=(4)/(30)`
`F=(30)/(4)` cm
object and image coincides `u=x=2f=15cm`
(b) `(1)/(f_(omega))=((4)/(3)-1)((1)/(infty)-(1)/(-60))=(1)/(180)`
Combines focal length of two lenses.
`(1)/(F')=(1)/(f_(omega))+(1)/(f_l)=(1)/(180)+(1)/(60)=(2)/(180)=(1)/(45)`
The image by combined lens should be formed at centre of curvature of mirror.
`(1)/(v)-(1)/(u)=(1)/(F')implies(1)/(-20)-(1)/(-x)=(1)/(45)`
`(1)/(x)=(1)/(20)+(1)/(45)`
`x'=(180)/(13)`
The distance by which pin should be shifted
`x=x'=15-(180)/(13)=(215-180)/(12)=(35)/(13)cm`
Alternatively:
`(1)/(F)=(1)/(f_(omega))+(1)/(f_l)+(1)/(f_m)+(1)/(f_(omega))`
`=(2)/(f_(omega))+(2)/(f_l)+(1)/(f_m)=(2)/(180)+(2)/(60)+(1)/((20)/(2))`
`=(1)/(90)+(1)/(30)+(1)/(10)=(+1+3+9)/(90)=(13)/(90)`
Since object and image coincides
`u=x'=2F=(180)/(13)`
Shift `=x-x'=15-(180)/(13)=(35)/(13)cm`