Question

What happens to the size of the image of the flame formed on the screen?

Answer 1

Given,

Distance between the object and lens$u=2m=200cm$

Focal length of the lens$f=10cm$

We know that,

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

$\Rightarrow \frac{1}{v}-\frac{1}{-200}=\frac{1}{10}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{200}$

$\Rightarrow v=\frac{200}{19}=10.52cm$

Now, Distance between object and screent$\left(d\right)=|u|+|v|=200+10.52=210.52cm$

Now, $u=d-v$

We know that,

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

$\frac{1}{v}=\frac{1}{f}+\frac{1}{d-v}(\because u=d-v)$

$\frac{1}{v}=\frac{d-v+f}{f(d-v)}$

$v^2-dv+df=0$

So, after solving,

$v=\frac{d\pm \sqrt{d^2-4df}}{2}$

We know that,

$m=-\frac{v}{u}$

So, there is two condition,

The lens can be moved towards the object to form an image on the screen, In this case, inverted real image is enlarged. The image is closed to the focal point. So when the value of $u$ decreases, then the value of $v$ will be increased.

So the size of the image will gradually increase, as $u$ will decrease.