Question

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

Answer 1

Hint: First use the lens formula, i.e. $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ to calculate the value of $u$. Here ${h_o} = 2cm$. Then use the equation $m = - \dfrac{v}{u} = \dfrac{{{h_i}}}{{{h_o}}}$ to calculate the value of ${h_i}$. The size of the picture will be ${h_i} \times {h_i}cm$. For the second part let $P$ be the power of the source. The illumination of both the slide and the screen will be given by the equation, illumination $ = \dfrac{P}{{Area}}$, and then calculate the ratio of the illumination of both the slide and the screen.

Complete answer:
In the problem, we are given
The focal length of the projector lens, $f = 10cm$
The distance of the image from the lens, $v = 5m = 500cm$
Height of the object, ${h_o} = 2cm$

A.)Let, the distance of the object from the lens be $u$ and the height of the image be ${h_i}$
We know, by the lens formula that
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$ \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 u = - \dfrac{{500}}{{49}}cm$
We also know that the equation for linear magnification is
$m = - \dfrac{v}{u} = \dfrac{{{h_i}}}{{{h_o}}}$
$ \Rightarrow - \dfrac{{v \times {h_o}}}{u}$
$ \Rightarrow {h_i} = - \dfrac{{500 \times 2}}{{\dfrac{{500}}{{49}}}}$
$ \Rightarrow {h_i} = - 98cm$
Negative size indicates that the image is inverted.
Hence, the size (dimensions) of the image on the screen is $98cm \times 98cm$ .

B.)Let $P$ be the illuminating power of the source which will be divided equally all over the area of the slide and the picture on the screen.
Illumination on any surface is given by
Illumination $ = \dfrac{P}{{Area}}$
Let illumination on the slide be ${I_1}$ and the illumination on the screen be ${I_2}$ .
So, ${I_1} = \dfrac{P}{{2 \times 2}}$
And ${I_2} = \dfrac{P}{{98 \times 98}}$
So, the ratio of the illumination on the slide to the illumination of the screen is equal to
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\dfrac{P}{{2 \times 2}}}}{{\dfrac{P}{{98 \times 98}}}}$
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{49 \times 49}}{1} = \dfrac{{2401}}{1}$
Hence, the ratio of the illumination on the slide to the illumination of the screen is $\dfrac{{2401}}{1}$ .

Note:
In the solution of the first part we see that the image of the slide on the screen is larger than the size of the screen, the only lens that can produce a real image larger than the size of the object is convex. This is why convex lenses are used in projectors to display the image of a small object on a larger screen.