Question

(i) An object is placed at 5 om trom a convex lens. If focal length of convex lensis 20 cm. then
find the position and nature of the image?​

Answer 1

Given :-

An object is placed at 5cm from a convex lens and the focal length of the lens is 20cm .

To Find :-

The position and the nature of the image = ?

Used Concepts :-

• Lens Formula i.e 1/v - 1/u = 1/f , where v = image distance , u = object distance and f = focal length of the spherical lens respectively .
• object distance of both spherical lenses is ( -ve )
• The ( +ve ) sign of the image distance shows us that the image formed is real and inverted .
• Focal length of a convex lens is positive and that of a concave lens is negative .

Solution :-

Here , object distance ( u ) = - 5cm

Focal length ( f ) = 20cm

By lens formulae ,

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

Putting all the values ,

$\frac{1}{v} = \frac{1}{20} + \frac{1}{( - 5)}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{5}$

$\frac{1}{v} = \frac{1 - 4}{2 0}$

$\frac{1}{v} = \frac{ - 3}{20}$

Therefore , v = -20/3 cm

Now , the negative sign of the image distance shows us that the image is formed as on the same side of the object . And we knows that if any image is formed on the same side of the object the image is erect and virtual .

Therefore , The image is formed -20/3 cm in front of the lens and the nature of the image is virtual and erect.

Answer 2

Given:-

An object is placed at 5 om trom a convex lens. If focal length of convex lensis 20 cm.

To find:-

find the position and nature of the image?

$\huge\color{cyan}\boxed{\colorbox{black}{⚛solution:-}}$

Given :--

U-20

F-5

V-??

Answer by using lens formula :-

1/f=1/v-1/u

1/5=1/v-(-1)/5

1/v=2-1/5

v=5 (Answer)

Now , We have to find magnification

Magnification = -v/u

=5/20

=-4

Nature = Real

___________________________________