Question

A $35$mm film is to be projected on a $20$m wide screen situated at a distance of $40$m from the film projector. Calculate the distance of the film from the projection lens and focal length of the projection lens.
A. $7\;cm,\;\;70\;mm$
B. $6\;cm,\;\;60\;mm$
C. $5\;cm,\;\;50\;mm$
D. $4\;cm,\;\;40\;mm$

Answer 1

As in case of projector, lenses are used. To find the distance of film from a projection lens, magnification of the lens is to be used and for focal length, lens formula can be used.

Formula used:
(i) Magnification of lens $= \; + \dfrac{v}{u} = \dfrac{I}{O}$
(ii) Lens formula
$\dfrac{I}{v} - \dfrac{I}{u} = \dfrac{I}{f}$
Where u $=$ distance of object from the lens
v $=$ distance of image from the lens
f $=$ focal length of the lens

Complete step by step answer:
The film to be projected using a projector acts as the object and the lens in the projector will help in image formation that is for projecting the film on the screen.
Now, magnification is given by
$m = \dfrac{{size\;of\;image}}{{size\;of\;object}} = \dfrac{I}{O}$ .….(1)
Where I $=$ size of image
And O $=$ size of object
Also, in case of lenses, magnification is
$m = + \dfrac{v}{u}$ …..(2)
Where v is distance of image from lens and u is distance of object from lens.
So, from (1) and (2), we can say that
$\dfrac{I}{O} = + \dfrac{v}{u}$ ….(3)
As in this case,
Size of object, O $=$ Size of film $=$ $35\;mm$
$\Rightarrow$ $O = 35 \times {10^{ - 3}}m$
and size of Image, I $=$ Size of the projector
$\Rightarrow$ I$=$ $20\;m$
v $=$ distance between screen and the projector as image will form on screen.
U $=$ distance of film from the projection lens.
Putting all these values in equation (3), we get
$- \dfrac{{20\;m}}{{35 \times {{10}^{ - 3}}m}}\;\; = \;\; + \dfrac{{40\;m}}{u}$
$\Rightarrow$ $u = - \dfrac{{40 \times 35 \times {{10}^{ - 3}}}}{{20}}$
$\Rightarrow$ $u\; = \; - 2 \times 35 \times {10^{ - 3}}$
$\Rightarrow$ $u\; = \; - 70 \times {10^{ - 3}}m = - \,0.07\;m$
$\Rightarrow$ $u\; = \;7\;cm$
Now, the lens formula is given by
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\;\; \Rightarrow \;\;\dfrac{1}{{4000}} - \dfrac{1}{{ - 7}}\; = \;\dfrac{1}{{4000}} + \;\dfrac{1}{7}$
$\Rightarrow$ $\dfrac{1}{f}\;\; \Rightarrow \;\;\dfrac{{ + \,7\, + \,4000}}{{4000 \times 7}}\;\; \Rightarrow \;\;\;\dfrac{1}{f}\; = \;\dfrac{{4007}}{{(4000) \times 7}}$
$\Rightarrow$ $f = \;\dfrac{{4000 \times 7}}{{4007}}\;\; \Rightarrow \;\;f = 7cm\;(approx)\; = 70\;mm$
$\therefore$ the focal length of projection lens is $70\;mm$

So, the correct answer is “Option A”.

Note:
As u is the negative, this indicates that the object is placed in front of the projector lens. The values of I and O are in accordance with sign conventions.