Question

The minimum distance between a real object and its real and its real image formed by a thin converging lens of focal length f is

• A. $4f$
• B. $2f$
• C. $f$
• D. $f/2$

Answer 1

The correct option is A $4f$

From the figure it can be seen that

The object distance will be $u=-(x-v)$

Applying lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{-\left(x-v\right)}$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{\left(x-v\right)}\Rightarrow fx=vx-v^2$

$\therefore v^2-cx+fx=0$

Apply the quadratic equation for real roots $D⩾0$

$D=b^2-4ac⩾0$

$x^2-4xf⩾0\Rightarrow x^2⩾4xf$

$\therefore x⩾4f$

Hence the minimum distance between the real object and real image formed by a thin converging lens of focal length $f$ is $4f$ .