A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90cm.

The correct options are
B The distance of the object from the lens in its first position is 36cm.
C The height of the object is 6cm.
D The focal length of the lens is 21.6 cm.
For the two real images u and v get interchanged
$\Rightarrow \frac{v}{u}=\frac{9}{h_0}$ and $\frac{u}{v}=\frac{4}{h_0}$
$\frac{36}{h_0^2}=1\Rightarrow h_0=6cm$
Also v+u=90 and 6u=9v
$v=\frac{9v}{6}=90$
$\Rightarrow$ v=54 and u=36
The distance between the two lens positions is v - u = 28 cm
Also $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{54}-\frac{1}{-\left(36\right)}=\frac{1}{f}$
$\Rightarrow$ f=21.6 cm