A screen is placed 90 cm from an object . The image of the object on the screen is formed by convex lens at two different locations separated by 20 cm . Determine the focal length of the lens (b) suppose the object has a size of 10 cm, and in one case the image is half of its size. Determine image size when lens is displaced to second position

When the object and screen (where image forms) are D = 90cm apart, a

displacement of lens (x=20cm) produces image on screen.

By lens. Displacement method focal length :-

$f=\frac{D^2-x^2}{4D}=\frac{{90}^2-{20}^2}{2\times 90}\simeq 42.78cm.$

Now, height of object, h= 10cm. In one case, Image size is by = 5cm.

In other case the image size is $h_2$, which is related by :-

$h=\sqrt{h_1{h}_2}$

$\Rightarrow h_1=\frac{h^2}{h_1}=\frac{{10}^2}{5}=20cm$ (Ans)