Question

A projector lens has a focal length of 10cm. It throws an image of a 2cm x 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

Hint:The focal length and the distance of the image from the lens is given. Therefore use the lens formula to find the distance of the source from the lens. Now use the formula for linear magnification of an image to find the size of the picture on the screen.

Formula used:use the lens formula, i.e. $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ where f is the focal length, u is the distance of the object and v is the distance of the image.
Also, the formula for linear magnification of image is given by $m = - \dfrac{v}{u} = \dfrac{{h'}}{h}$

Complete step-by-step solution:
The lens of a projector has a focal length of 10 cm. It gives an image of a given 2cm×2cm slide on a screen which is 5 meters away from the lens. 
 f = 10cm
Distance of the image from the lens v = 5 m = 500 cm
Height of the object, h ​= 2 cm
We know, the lens formula is as follows.
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Applying this formula, we get

$ \Rightarrow \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}$
$ \Rightarrow \dfrac{1}{u} = \dfrac{1}{{500}} - \dfrac{1}{{10}}$
$ \Rightarrow \dfrac{1}{u} = \dfrac{{1 - 50}}{{500}}$
$ \Rightarrow \dfrac{1}{u} = - \dfrac{{49}}{{500}}$
$ \Rightarrow u = - \dfrac{{500}}{{49}}$
Therefore, the value of linear magnification is,
$m = - \dfrac{v}{u} = \dfrac{{h'}}{h}$
$ \Rightarrow h' = - \dfrac{{v \times h}}{u}$
$ \Rightarrow h' = - \dfrac{{500 \times 2}}{{ - \dfrac{{500}}{{49}}}}$
$ \Rightarrow h' = 98{\text{ cm}}$
Therefore the size of the image on screen is 98x98 cm.
(ii) Let the illuminating power of the source be denoted by P.
 Therefore, the illumination of the slide is = $\dfrac{p}{{2 \times 2}}$ ​
The illumination of the image is given by =$\dfrac{p}{{98 \times 98}}$
 ​So the ratio of illuminations of the slide and of the image on the screen is $\dfrac{{\dfrac{p}{{2 \times 2}}}}{{\dfrac{p}{{98 \times 98}}}} = 49 \times 49:1 = 2401:1$

Note:We know that the lens formula is $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ , where f is the focal length, u is the distance of the object and v is the distance of the image.
Also, the formula for linear magnification of image is given by $m = - \dfrac{v}{u} = \dfrac{{h'}}{h}$
Note that, negative sign before the magnitude of height denotes that the image is inverted.