The following figure shows a beam of light converging at point P. When a concave lens of focal length 16 cm is introduced in the path of the beam at a place shown by dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value of x will be equal to

Question

The following figure shows a beam of light converging at point P. When a concave lens of focal length 16 cm is introduced in the path of the beam at a place shown by dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value of x will be equal to (A) 12 cm (B) 24 cm (C) 36 cm (D) 48 cm

Tardigrade Admin · Accepted Answer

So, here when we put the concave lens,
let the beam will converge at a distance

x = v

Using lens formulae, we have,

1/ f = 1/ v - 1/ u

Where

u = 12 c m

and

f = - 16 c m

is given

∴ 1/ v = (1/ f) + (1/ u)

= (- 1/16) + (1/12) = 1/48 c m

\Rightarrow v = 48 c m

Hence,

x = 48 c m

12 cm

24 cm

36 cm

48 cm

So, here when we put the concave lens, let the beam will converge at a distance x=v Using lens formulae, we have, 1/f=1/v−1/u Where u=12cm and f=−16cm is given ∴1/v=(1/f)+(1/u) =(−1/16)+(1/12)=1/48cm ⇒v=48cm Hence, x=48cm

So, here when we put the concave lens,
let the beam will converge at a distance $x = v$
Using lens formulae, we have, $1/ f = 1/ v - 1/ u$
Where $u = 12 c m$ and $f = - 16 c m$ is given
$∴ 1/ v = (1/ f) + (1/ u)$
$= (- 1/16) + (1/12) = 1/48 c m$
$\Rightarrow v = 48 c m$
Hence, $x = 48 c m$