Question

Rohit focused the image of a candle flame on a white screen using a convex lens.
He noted down the position of the candle, screen, and lens as under:
Position of candle = $26.0$cm
Position of convex lens = $50.0$ cm
Position of screen = $74.0$ cm

What is the focal length of the convex lens?
Where will the image be formed if he shifts the candle towards the lens at a position of $38$ cm?
Draw a ray diagram to show the formation of the image in case (ii) as said above?

Answer 1

Position of candle$=26$cm

Position of convex lens$=50$cm

Position of screen$=74$cm

(i) Candle distance from lens, $v=50-26=24$cm

image distance from lens(on screen), $v=74-50=24$

then, $v=u$

It means object is placed at centre of curvature of lens.

So, $v=2f\Rightarrow f=\frac{v}{2}=\frac{24}{2}$cm

$\therefore$ (focal length) $f=12$cm

(ii) Position of candle$=38$cm

Object distance from lens$=50-38=12$cm

$\therefore u=-12$cm, $f=12$cm

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

$\frac{1}{12}=\frac{1}{v}-\frac{1}{(-12)}$

$\frac{1}{v}=\frac{1}{12}-\frac{1}{12}$

$v=\infty$.