Question

an object 5 cm in length is held 25 cm away form a convex lens of focal length 10 cm find position, size and nature of image​

Answer 1

Given :

In convex lens,

Object height = 5cm

Object distance = - 25cm

Focal length = 10cm

To find :

The position, size and nature of image

Solution :

Using lens formula,

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 :

$\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}$

here,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

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

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{ - 25}= \dfrac{1}{10}}$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{ 25}= \dfrac{1}{10}}$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{1}{ 10} - \dfrac{1}{25}}$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{5 - 2}{ 50} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{3}{ 50} }$

$\leadsto{ \sf v = \dfrac{50}{3} }$

$\leadsto{ \sf v = 16.6 \: cm }$

thus, position of image is 16.6cm.

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

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

here,

• m denotes magnification
• h' denotes image height
• h denotes object height
• u denotes object distance
• v denotes image distance

by substituting all the given values,

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

$\leadsto{ \sf \dfrac{h'}{5} = \dfrac{16.6}{25}}$

$\leadsto{ \sf h'= \dfrac{16.6 \times 5}{25}}$

$\leadsto{ \sf h'= \dfrac{83}{25}}$

$\leadsto{ \sf h'= 3.32 \: cm}$

thus, the size of the image is 3.32cm.

Nature of image :

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