Question

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 90 cm.

Answer 1

$h_{object}^2=h_{image1}\times h_{image2}$

$h_{object}=\sqrt{36}=6$

magnification of image is $\frac{v}{u}=\frac{9}{6}$

$v=\frac{3u}{2}$

in lens displacement method , $u+v=d$ ; $uv=df$

$u+\frac{3u}{2}=90$

$u=36$ => $v=54$

$uv=df$

$f=\frac{36\times 54}{90}=21.6$

option $D$ is correct