Question

Q33. An object of height 5 cm is held 30 cm away from a converging lens of focal length
10 cm. Find the position, size and nature of the image formed.3​

Answer 1

Given :

In convex lens,

Object height = 5cm.

Object distance = - 30cm

Focal length = 10cm.

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,

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

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{( - 30)}= \dfrac{1}{10}}$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{30}= \dfrac{1}{10}}$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{30}}$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{3 - 1}{30} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{1}{15} }$

$\leadsto{ \sf v = 15 \: cm }$

thus, the position of image is 15cm.

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,

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

$\leadsto{ \sf \dfrac{h'}{5} = \dfrac{15}{ - 30} }$

$\leadsto{ \sf {h'} = - \dfrac{15 \times 5}{30} }$

$\leadsto{ \sf {h'} = - \dfrac{75}{30} }$

$\leadsto{ \sf {h'} = 2.5 \: cm}$

thus, the size of image is 2.5cm

Nature of image :

• The image is real and inverted.
• The image is diminished.