Question

In which direction does he move the lens to focus the flame 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,

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

So, there is two condition,

$\text{Condition 1:}$ The lens can be moved towards 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 value of $u$ decreases, then value of $v$ will be increased.

$\text{Condition 2:}$ In this case lens is moved towards the screen, In this case image distance is less than initial. So when $u$ increases, then $v$ decreases.