Question

The distance between an object and the screen is $75cm$ . When the convex length of focal length $12cm$ is placed in between the object and screen , magnification of real image formed can be :

• A. $4$
• B. $2$
• C. $8$
• D. $16$

Answer 1

The correct option is A $4$

It is given that distance between and object and screen is $75cm$.

So we can write $v=75-u$

Applying the lens formula

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

$\frac{1}{12}=\frac{1}{75-u}-\frac{1}{-u}=\frac{75}{75u-u^2}$

$900=75u-u^2$

$u^2-75u+900=0$

$\therefore u=\frac{75\pm \sqrt{{75}^2-4\times 900}}{2}$

$\Rightarrow u=\frac{75\pm 45}{2}=\frac{120}{2},\frac{30}{2}$

$\Rightarrow u=60cm,15cm$

For $u=15cm$

$\frac{1}{v}-\frac{1}{-15}=\frac{1}{12}$

$\Rightarrow \frac{1}{v}=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}=\frac{1}{60}$

$\therefore v=60cm$

Hence magnification, $m=-\frac{v}{u}=-\frac{60}{-15}=4$

Hence the correct answer is option (A).