Question

The minimum distance between an object and its real image formed by a convex lens is

Answer 1

Step 1: Formula used

We know, the Lens formula is $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

where,

$v\to$ distance of image from optical center

$u\to$ distance of object from optical center

$f\to$ focal length of the lens

Step 2: Assume the distances and substitute them in lens formula

Let us assume,

Distance between the object and real image formed be $x$

Object distance, $u=-y$

Image distance, $v=x-y$

So, the lens equation becomes,

$\frac{1}{x-y}-\frac{1}{-y}=\frac{1}{f}$

$\Rightarrow \frac{1}{x-y}+\frac{1}{y}=\frac{1}{f}$

$\Rightarrow y^2-xy+fx=0$

Step 3: From the quadratic equation, calculate the min value of

Value of $y$,

$y=\frac{-x\pm \sqrt{x^2-4fx}}{2}$

For $y$ to be real, $-x\pm \sqrt{x^2-4fx}\ge 0$

$\Rightarrow x^2-4fx\ge 0$

$\Rightarrow x^2\ge 4fx$

$\Rightarrow x\ge 4f$

$\because$ The minimum distance between an object and its real image formed by a convex lens is $4f$.