A convex lens when placed in the first position forms a real image of an object on a fixed screen. The distance between the object and the screen is $75\text{cm}$. On displacing the lens from first position by $25\text{cm}$ to the second position, again a real image is formed on the screen. Then the focal length of the lens is

The correct option is B $\frac{50}{3}\text{cm}$
From the lens displacement method, we can write that,
$f=\frac{D^2-x^2}{4D}...\left(1\right)$

where, $f\to$ Focal length
$D\to$ distance between screen,
$x\to$ distance between the two position of lenses,

Substituting the data given in the question in equation $\left(1\right)$, we have

$f=\frac{(75{)}^2-(25{)}^2}{4\times 75}$

$\Rightarrow f=\frac{5000}{300}$

$\therefore f=\frac{50}{3}\text{cm}$

Hence, option (b) is the correct answer.