Question

A 10cm tall object is placed perpendicular to the principal axis of convex lenses of focal length 30cm. The distance of the object from the lense is 20cm. Find the: 1) Position, 2) Nature, 3) Size of the image formed.

Answer 1

Given :

In convex lens

Object height = 10cm.

Focal length = 30cm.

Object distance = -20cm.

To find :

(a) Position of the image

(b) size of the image formed

(c) Nature of the image

Solution :

(a) Using lens formula,

The relationship between image distance, object distance and focal length of a lens is known as 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

Now, substituting all the given values,

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

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{ (- 20)} = \dfrac{1}{30} }$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{20} = \dfrac{1}{30} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{1}{30} - \dfrac{1}{20} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{3 - 2}{60} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{1}{60} }$

$\leadsto{ \sf v = 60 \: cm}$

thus, the image distance is 60cm.

(b) Magnification

The Magnification produced by a lens is equal to the ratio of image distance to the object distance .i.e.,

$\leadsto{ \bf m = \dfrac{v}{u }}$

$\leadsto{ \sf m = \dfrac{60}{ - 20 }}$

$\leadsto{ \bf m = - 3}$

we know,

$\leadsto{ \bf m = \dfrac{h'}{h}}$

$\leadsto{ \sf - 3 = \dfrac{h'}{10}}$

$\leadsto{ \sf {h'} = - 3 \times 10}$

$\leadsto{ \sf {h'} = - 30 \: cm}$

thus, size of the image is 30cm.

(c) Nature of image :

• The image is real and inverted
• The image is enlarged