Show that the least possible distance between an object and its real image in a convex lens is 4f where f is the focal length of the lens.

Solution

To prove: Least possible distance between an object and real image = 4f

Suppose, I is the real image of an object O.
Let, d be the distance between the image and the object.
If the image distance from the lens is x, the object distance from the lens will be (d – x).

Thus, u = – (d – x) and v = + x.

Sustituting in the lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
we have,
$\frac{1}{x} - \frac{1}{- (d - x)} = \frac{1}{f}$

$\Rightarrow$ $\frac{1}{x} + \frac{1}{(d - x)} = \frac{1}{f}$

$\Rightarrow$ $x^{2} - xd - fd = 0$

For a real image, the value of x must be real,
i.e., the roots of the above equation must be real. This is possible if,
d² ≥ 4fd
i.e., d ≥ 4f

Hence, 4f is the minimum distance between the object and its real image formed by a convex lens.

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Wave Optics

Show that the least possible distance between an object and its real image in a convex lens is 4f where f is the focal length of the lens.

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