The convex surface of a thin concavo-convex lens of glass of R.l. 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 horizontally.
Q If the concave part is filled with water of refractive index 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

In this case, the effective focal length F. of the combination is given as
`- 1/F^(.) = 1/f_(w) + 1/f_(g) - 1/f_(m) + 1/f_(g) + 1/f_(w)`
`- 1/F^(.) = 2/f_(w) + 2/f_(g) - 1/f_(m)`, where `1/f_(w) = (n-1)(1/R_(2)) = (4/3 -1) (1/60)= 1/180`
`implies f_(w)=180cm`
Putting the values of `f_(w)= 180cm, f_(g)=60cm` and `f_(m)= -10cm`
We obtain `f_(1)= + 60cm`.
`-1/f^(.) = 2/180 + 2/60 - 1/(-10) implies f^(.) = -(90/13) cm`
If the object distance is `l^(.)` for the object to coincide with the image,
`l^(.) = 2F^(.) = 180/13 cm implies` The displacement of the object.
`= l - l^(.) = 15 - 180/13 = 15/13 = 1.15 cm`