Question

An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Calculate the position and magnification of the image and state its nature. 

Answer 1

Distance of the object from the mirror 'u' = -5 cm
Focal length of the mirror 'f '= 10 cm
We have to find the distance of the image 'v'.
Using the mirror formula, we get 

${\displaystyle \frac{1}{f}=\frac{1}{v}+\frac{1}{u}}$ 

⇒${\displaystyle \frac{1}{10}=\frac{1}{v}+\frac{1}{-5}}$ 

⇒${\displaystyle \frac{1}{10}=\frac{1}{v}-\frac{1}{5}}$ 

⇒${\displaystyle \frac{1}{10}+\frac{1}{5}=\frac{1}{v}}$ 

⇒${\displaystyle \frac{1}{10}+\frac{2}{10}=\frac{1}{v}}$ 

⇒${\displaystyle \frac{1}{v}=\frac{3}{10}}$ 

⇒${\displaystyle v=\frac{10}{3}}$ 

⇒${\displaystyle v=3.3}$ cm 

Thus, the distance of the image 'v' is 3.3 cm behind the mirror.Now, using the magnification formula, we get 

${\displaystyle m=\frac{-v}{u}}$ 

${\displaystyle m=\frac{-3.3}{-5}}$ 

m = 0.66 

Thus, the image is virtual, erect and small in size.