### Chapter 1: Light - Reflection and Refraction

Q
##### Light - Reflection and Reflection

Question:

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Focal length ($f$) = 15 cm

Object distance ($u$) = -10 cm

Image distance ($v$) = ?

Object distance ($u$) is taken negative as their distances from the pole of the mirror is measured opposite to the direction of the incident light.

And, focal length ($f$) is taken positive according to the sign convention, as the mirror is convex.

By using the mirror formula, we get

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

$\frac{1}{15} = \frac{1}{v}+\frac{1}{(-10)}$

$\frac{1}{15} +\frac{1}{10} = \frac{1}{v}$

$\frac{1}{v} = \frac{2+3}{30}$

$\frac{1}{v} = \frac{5}{30}$

$v = 6 cm$

The positive sign of $v$ indicates that the image is formed behind the convex mirror at a distance of 6 cm. That is the image formed is virtual in nature.

Now to find the nature of the image we will find the magnification.

$m= -\frac{v}{u}$

$m= -\frac{6}{-10} = 0.6$

The positive value of $m$ indicates that the image formed is erect and the value of $m$ is less than 1 indicates the image formed is diminished in size.

Therefore, the image formed at a distance of 6 cm behind the mirror. The image formed is virtual, erect and diminished in size.

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