Question

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

Answer 1

a) $f=10cm$, $V=5m=500cm$

$\frac{1}{V}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{500}-\frac{1}{u}=\frac{1}{10}$

$u=-\left[\frac{500}{49}\right]cm$

$m=\frac{V}{u}=\frac{500}{\left(\frac{-500}{49}\right)}=-49$

Here negative sign implies the image is inverted with respect to object. Now, as_here object is $(2cm\times 2cm)$ so the size of picture on the screen is

$A_1=(2\times 49cm)\times (2\times 49cm)$

$=(98\times 98){cm}^2$

b) As light energy passing per sec through slide is equal to that in the picture on the screen.

$IA=I_1{A}_1$

$\frac{I}{A}=\frac{A_1}{A}=\frac{ma\times mb}{a\times b}=m^2$

$={(-49)}^2=49\times 49$

i.e; intensity in picture on the screen will be $\left[\frac{1}{49\times 49}\right]$ times lesser than that on the slide. This is why in case of projector for observing image on a screen the source of light must be very powerful and the room dark.