A real image is formed by a convex lens. If we put a concave lens in contact with it, the combination again forms a real image. The new image

Options

A. nearer to the lens system
B. at the original position
C. at a larger distance from the lens system
D. anywhere depending on the focal length of the concave lens

Answer

C

Explanation

When a concave lens is kept in contact with the convex lens, the effective power of the lens `(P_(1)-P_(2))` is decreased and since `P=(1)/(f)`, the focal of the combined lens (F) is increased, but still it acts as a convex lens.
In both cases real images are formed but `v_(1)ne v_(2)`
`because F gt f` and u is the same and it is negative.
In the first case,
`(1)/(v_(1))-(1)/(u)=(1)/(f) or (1)/(v_(1))=(1)/(f)+(1)/(u)=(1)/(f)-(1)/(u)`
In the second case, `(1)/(v_(2))=(1)/(F)-(1)/(u)`
But `because F gt f, v_(2) gt v_(1)`
Thus the image is formed at a larger distance.
[e.g. Suppose that f=20, F=30 and u=50
Then, `(1)/(v_(1))=(1)/(20)-(1)/(50)=(30)/(1000) therefore v_(1)=(100)/(3)`
and `(1)/(v_(2))=(1)/(30)-(1)/(50)=(20)/(1500) therefore v_(2)=75`
Thus `v_(2) gt v_(1)`]