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

Asked by pratikbharadia | 10th Dec, 2009, 08:33: AM

Expert Answer:

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

Answered by  | 10th Dec, 2009, 04:32: PM

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