Question

an object of height 5cm kept at a distance of 20cm from the optic of concave lens of focal length 8cm find the position and height of the image​

Answer 1

Given :

In concave lens,

Object height = 5 cm

Object distance = - 20 cm

Focal length = -8 cm

Remember : In concave lens focal length is negative.

To find :

The position and height of the iimage

Solution :

Using lens formula,

A 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,

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

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{ - 20} = - \dfrac{1}{8} }$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{20} = - \dfrac{1}{8} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{8} - \dfrac{1}{20} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{ - 5 - 2}{40} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{ - 7}{40} }$

$\leadsto{ \sf v = - \dfrac{40}{7} }$

$\leadsto{ \sf v = - 5.7 \: cm}$

thus, position of image is -5.7 cm.

Now, using Magnification that is

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

here,

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

by substituting all the given values,

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

$\leadsto{ \sf \dfrac{h'}{5}= \dfrac{5.7}{20}}$

$\leadsto{ \sf h'= \dfrac{5.7 \times 5}{20}}$

$\leadsto{ \sf h'= \dfrac{28.5}{20}}$

$\leadsto{ \sf h'= 1.42 \: cm}$

thus, the image height is 1.42cm.