Practicing Success
Practicing Success
CUET Preparation Today
The minimum distance between real object and real image in the context of convex lens of focal length'f' is |
2f f $\infty$ 4f |
4f |
Let the distance between object and its real image is d. Let the distance of object from the lens is x
So the distance of image and lens be d−x
Using the lens equation :
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$ \frac{1}{f} = \frac{1}{d-x} - \frac{-x} = \frac{d}{(d-x)x}$
$ x^2 -dx + fd = 0 $
$\text{For x to be real } d^2 -4df \ge 0$
$ \Rightarrow d \ge 4f$ |
