Question

In a convex lens of focal length F, the minimum distance between an object and its real image must be :

• A. 3F
• B. 4F
• C. 5F
• D. 6F

Answer 1

The correct option is B 4F

Let image distance be $v$, object distance be $-u$ and focal length be $f$.

From lens formula, $\frac{1}{v}-\frac{1}{-u}=\frac{1}{f}$

Let distance between object and image be $s$,

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

$\frac{s}{u(s-u)}=\frac{1}{f}$

$s=\frac{u^2}{u-f}$

For point of minima,

$\frac{ds}{du}=0$

$\frac{u^2-2uf}{(u-f{)}^2}=0$

$u=0,2f$

$s=\frac{(2f{)}^2}{2f-f}=4f$