Question

A 2 cm long object is placed at a clistance of 100 cm
from a convex mirror of radius of curvature 50 cm.
Find the position, nature and length of the image.
Ans.20 cm behind the mirror, virtual, 0.4 cm.​

Answer 1

Given :

$\bullet\;\sf{height\:of\:object\:,\:h_o=2\;cm}$

$\bullet\;\sf{position\:of\:object\:,u\;=-100\;cm}$

$\bullet\;\sf{Radius\:of\:curvature\:of\:convex\:mirror\;,R=50\;cm}$

To find :

$\bullet\;\sf{position\:of\:image\:,v\;=?}$

$\bullet\;\sf{nature\:of\:image}$

$\bullet\;\sf{height\:of\:image,\;h_i\;=?}$

Formula required :

• Relation between Radius of curvature and focal length of Mirror

$\red{\bigstar}\boxed{\sf{f=\dfrac{R}{2}}}$

• Formula for magnification of mirror

$\red{\bigstar}\boxed{\sf{m=\dfrac{-v}{u}=\dfrac{h_i}{h_o}}}$

• Mirror formula

$\red{\bigstar}\boxed{\sf{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}$

[ where m is magnification ; v is position of image ; u is position of object ; f is focal length of mirror ; R is radius of curvature of convex mirror ; h$\sf{_i}$ is height of image ; h$\sf{ _o}$ is height of object ]

Solution :

Using relation between R and f

$\implies\sf{f=\dfrac{R}{2}}$

$\implies\sf{f=\dfrac{50}{2}}$

$\underline{\underline{\implies\sf{\red{f=25\;cm}}}}$

Using mirror formula

$\implies\sf{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}$

$\implies\sf{\dfrac{1}{25}=\dfrac{1}{v}+\dfrac{1}{(-100)}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{1}{25}+\dfrac{1}{100}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{4+1}{100}}$

$\underline{\underline{\implies\sf{\red{v=20\;cm}}}}$

Using formula for magnification

$\implies\sf{m=\dfrac{-v}{u}=\dfrac{h_i}{h_o}}$

$\implies\sf{\dfrac{-v}{u}=\dfrac{h_i}{h_o}}$

$\implies\sf{\dfrac{-(20)}{(-100)}=\dfrac{h_i}{(2)}}$

$\underline{\underline{\implies\sf{\red{h_i=0.4\;cm}}}}$

Therefore,

• position of image is 20 cm behind the convex mirror.
• nature of image is virtual and erect.
• Height of image is 0.4 cm.