Question

an object of height 4 cm is placed 16 cm away from the optical centre of a convex lens of focal length 24 cm find the position size and nature of the image formed

Answer 1

Given :

In convex lens,

Object height = 4cm.

Object distance = - 16cm

Focal length = 24cm.

To find :

The position, size and nature of the image formed.

Solution :

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,

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

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{( - 16)}= \dfrac{1}{24}}$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{16}= \dfrac{1}{24}}$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{1}{24} - \dfrac{1}{16}}$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{2 - 3}{48} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{48} }$

$\leadsto{ \sf v = - 48\: cm }$

thus, the position of image is -48cm.

Now, to find size of image we have to use magnification formula that is

$\boxed{ \bf m = \dfrac{h'}{h} = \dfrac{v}{u} }$

where,

• m denotes magnification
• h' denotes image height
• h denotes object height

by substituting all the given values in the formula,

$\leadsto{ \sf \dfrac{h'}{h} = \dfrac{v}{u} }$

$\leadsto{ \sf \dfrac{h'}{4} = \dfrac{-48}{ -16} }$

$\leadsto{ \sf {h'} = \dfrac{48 \times 4}{16} }$

$\leadsto{ \sf {h'} = \dfrac{192}{16} }$

$\leadsto{ \sf {h'} = 12\: cm}$

thus, the size of image is 2.5cm

Nature of image :

• The image is erect and virtual.
• The image is enlarged.