Question

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20cm,and (b) a concave lens of focal length 16cm.​

Answer 1

Given :

Object distance = 12cm

Focal length in convex lens = 20cm

Focal length in concave lens = 16cm

To find :

Position of the image in concave lens and convex lens.

Solution :

Here the object is virtual and the image formed will be real.

(a) Using lens formula that is,

» The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}$

where,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{12}= \dfrac{1}{20}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{12}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{3 + 5}{60}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{8}{60}$

$\dashrightarrow\sf v = \dfrac{60}{8}$

$\dashrightarrow\sf v = 7.5 \: cm$

Thus, the image is formed 7.5cm away from the lens, towards it's right.

(b) In concave lens focal length is negative,

so, focal length = -16cm

Using lens formula,

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{12}= - \dfrac{1}{16}$

$\dashrightarrow\sf \dfrac{1}{v} = - \dfrac{1}{16} + \dfrac{1}{12}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 3 + 4}{48}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{48}$

$\dashrightarrow\sf v = 48 \: cm$

Thus, the image is formed 48cm away from the lens towards it's right.