Question

(a) List four characteristics of image formed by concave lens of focal length $20\,cm$ when the object is placed at a distance of $40\,cm$ from its optical center
(b) The size of the image formed of an object with a convex lens of focal length $20\,cm$ is observed to be reduced to $\dfrac{1}{3}$ of its size. Find the distance of the object from the optical center of the lens.

Answer 1

Hint: In order to solve this question we need to understand the definition of concave and convex lens. Concave lens is a diverging lens whose edge is thicker than its center and it diverges a parallel beam falling on it such that it appears to come from one point known as focal point on the same side of lens. Convex lens is a converging lens whose edge is thinner than its center and it converges a parallel beam falling on it on a point known as focal point on the other side of the lens.

Complete step by step answer:
(a) According to given problem focal length of concave lens $f = - 20\,cm$ and
Object distance $u = - 40\,cm$
Let image distance be $v$ the using lens formula we get
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Putting formula we get
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{20}} + \dfrac{1}{{40}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{{ - 2 + 1}}{{40}} \\
\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{40}}c{m^{ - 1}}$
$\therefore v = - 40\,cm$

Negative value of image size indicates that the image formed on the same side of the lens is virtual and it forms on the same point as the object.
Also magnification of image is equals to $m = \dfrac{v}{u}$
$m = \dfrac{{ - 40}}{{ - 40}} = 1$
Since magnification is $1$ so it means image size is same as object size and it is positive means image is erect in nature. So the four characteristics about image formed is
(i) Image formed is virtual in nature.
(ii) Image forms in the same location as the object.
(iii) Image size is the same as that of an object.
(iv) Image is erect in nature.

(b) Focal length of convex lens is $f = 20\,cm$ and image size is $\dfrac{1}{3}$ of object means
Magnification is $m = - \dfrac{1}{3}$
Hence by using magnification formula we get $m = \dfrac{v}{u}$
$ \Rightarrow u = - 3v$
Now using lens formula we get $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Now using $v = - \dfrac{1}{3}u$ and lens formula we get
$\dfrac{1}{{20}} = - \dfrac{3}{u} + \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{{20}} = - \dfrac{2}{u}$
$ \therefore u = - 2 \times 20 = - 40\,cm$

So the object distance is $40\,cm$ on the same side of the lens.

Note:It should be remembered that focal length of convex lens is positive as it is a converging lens so image forms on opposite sides of lens whereas focal length of concave lens is negative as it is a diverging lens so image forms on the same side of lens. Also if magnification is positive means image is erect in nature and if negative then image formed is inverted in nature.