Question

A convex lens forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of the object. If the increase in distance of the object is 6 cm, what is the shift of the screen? [8 Marks]

Answer 1

Let object and image distance be u and v respectively and focal length = f.
For the first case when magnification $m=\pm 3=\frac{v}{u}$
$\Rightarrow v=+3u\because$ image is real.
Using lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{3u}+\frac{1}{u}=\frac{1}{f}$
or $\frac{4}{3u}=\frac{1}{f}$ ….. (1) [2 Marks]

For the case when object distance is increased by 6 and $m=\pm 2=\frac{v}{u+6}$
$\Rightarrow v=2(u+6)$
Using lens formula again,
$\frac{1}{2(6+u)}+\frac{1}{6+u}=\frac{1}{f}$
or $\frac{3}{2(6+u)}=\frac{1}{f}$ ….. (2) [2 Marks]
Eqs. (1) and (2) gives
$\frac{4}{3u}=\frac{3}{2(6+u)}$
$\therefore 9u=48+8u$
$u=48cm$ [2 Marks]
Shift of screen = 3u – 2(6+u)
$=3\times 48–2(6+48)$
= 36 cm [2 Marks]