STD 12 - 9. Ray optics abd optical instruments — NEET STD 11 Science — Question

Let the distance of object from the lens is $x$

So the distance of image and lens be $d - x$

Using the lens equation :

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

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

$x =\frac{ d \pm \sqrt{ d ^2-4 fd }}{2}$

So for real $x$

$\sqrt{ d ^2-4 fd } \geq 0$

So $d \geq 4 f$

So minimum distance between object and image is $4 f$.