Question

A convex mirror has a radius of curvature of 22 cm. An object is placed 14 cm from the mirror. What is the position and nature of the image?​

Answer 1

⇒ Given:

Radius of curvature of given convex mirror = 22 cm

Distance of the object from the mirror = 14 cm

⇒ To Find:

The position and nature of the image formed.

⇒ Formula to be used:

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

⇒ Solution:

Radius of curvature = 22 cm

Focal length of the mirror:

$\sf{\dfrac{R}{2}=\dfrac{-22}{2}=11\:cm}$

This is because R = 2F.

Distance of the object from the mirror, u = -14 cm

The distance of the object from the mirror must be taken in the negative form as the mirror is a convex one.

Applying the mirror formula:

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

$\sf{\dfrac{-1}{11}=\dfrac{1}{v}+\dfrac{1}{-14}}$

$\sf{\dfrac{-1}{11}=\dfrac{1}{v}-\dfrac{1}{14}}$

$\sf{\dfrac{-1}{11}+\dfrac{1}{14}=\dfrac{1}{v}}$

$\sf{\dfrac{-14+11}{154}=\dfrac{1}{v}}$

$\sf{\dfrac{-3}{154}=\dfrac{1}{v}$

$\sf{v=\dfrac{-154}{3}}$

$\sf{v=-51.3\:cm}$

This means that the image is formed from a distance of 51.3 cm in front of the mirror.

Nature of the image formed:

 ✳ Virtual

 ✳ Slightly diminished than that of the object