Question

A projector lens has a focal length 10 cm. It throws an image of a $2cm\times 2cm$ Slide on a screen 5 meter away from the lens. Find
(i) The size of the picture on the screen and
(ii)ratio of illuminations of the slide and of the picture on the screen.

Answer 1

Given: A projector lens has a focal length 10 cm. It throws an image of a $2cm\times 2cm$ slide on a screen 5 meter away from the lens.

To find

(i) The size of the picture on the screen and

(ii)ratio of illuminations of the slide and of the picture on the screen.

Solution:

(i) As per the given criteria,

focal length of the lens, $f=10cm$

image distance, $v=5m=500cm$

Height of the object, $h_o=2cm$

Applying the lens formula, we get

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}⟹\frac{1}{u}=\frac{1}{v}-\frac{1}{f}⟹\frac{1}{u}=\frac{1}{500}-\frac{1}{10}⟹\frac{1}{u}=\frac{1-50}{500}⟹u=-\frac{500}{49}cm$

Therefore, the linear magnification becomes,

$m=-\frac{v}{u}=\frac{h_i}{h_o}⟹h_i=-\frac{v\times h_o}{u}⟹h_i=-\frac{500\times 2}{-\frac{500}{49}}⟹h_i=-98cm$

negative size indicates the image is inverted.

Hence the size of the picture on the screen is $98cm\times 98cm$

(ii)

Let P be the illuminating power of the source,

then illumination of slide = $\frac{P}{2\times 2}$

and illumination of the picture = $\frac{P}{98\times 98}$

So the required ratio of illuminations of the slide and of the picture on the screen

= $\frac{\frac{P}{2\times 2}}{\frac{P}{98\times 98}}=49\times 49=2401$