Question

An object is held at a distance of 50 cm from a diverging mirror of focal

length 25 cm. What is the position and size of the image formed?​

Answer 1

Explanation:

Given,

Size of the object \(h_{1}=3\ cm\)

Size of the image \(h_{2}=?\)

Object distance \(u=-50\ cm\),

Focal length \(f=25\ cm\)

Image distance \(v = ?\)

Using mirror formula \(\frac {1}{v}+\frac {1}{u}=\frac {1}{f}\)

\(\frac {1}{v}=\frac {1}{f}-\frac {1}{u}=\frac {1}{25}-\frac {1}{-50}=\frac {3}{50}\)

or \(v=\frac {50}{3}=16.67\ cm\)

As \(v\) is positive, image is behind the mirror. It is virtual and erect.

Now, using magnification formula \(m=\frac {h_{2}}{h_{1}}=-\frac {v}{u}\)

\(\frac {h_{2}}{3}=\frac {-50/3}{-50}=\frac {1}{3}\)

or size of the image is \(h_{2}=1\ cm\)