Show that the minimum distance between an object and its real image by a convex lens is 4f.

Hi 

Hear is your solution....

Let us have

u be the object distance,
s be the separation of object and image,
f be the focal length of the lens.

1 / u + 1 / (s - u) = 1 / f
f(s - u) + fu - u(s - u) = 0
f X s - us + u² = 0
u² = (u - f)s
s = u² / (u - f)

d X s / du = ( (u - f)(2u) - u² ) / (u - f) ²
= (u² - 2uf) / (u - f) ²
= u(u - 2f) / (u - f) ²

d X s / du = 0 when:
u(u - 2f) = 0
u = 0 or, for a real image, u = 2f.

When u < 2f, d X s / du < 0.
When u > 2f, d X s / du > 0.
u = 2f is therefore a minimum.

The object and image distances for minimum separation are both 2f, giving separation 4f.

Hope this satisfies your question...

Aruna