Question

9. 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 nature of the image formed.​

Answer 1

$\large{\underline{\boxed{ Answer}}}$

Object distance (u) = -5 cm

Focal length = -10 cm

image distance = ?

here, we use the mirror formula

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

now, put the given value,

$= > \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$= > \frac{1}{v} = \frac{ - 1}{10} - \frac{ - 1}{5}$

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

$= > \frac{1}{v} = \frac{ - 2 + 1}{10}$

$= > \frac{1}{v } = \frac{ - 1}{10}$

$= > v = - 10cm$

hence, image distance will be -10 cm.

The negative sign indicate that the image will formed in front of mirror.

The nature is virtual erect and diminished image.