Question

focal length of convex mirror is 40 cm and object height 10 cm is placed at a distance of 40 cm in front of the mirror find the distance of the image from the pole and size and nature of the image​

Answer 1

Focal length, f = 40 cm

Object height, ${h}_{o}$ = 10 cm

Object distance, u = -40cm

Image distance, v = ?

Image weight, ${h}_{i}$ = ?

Using mirror formula

$\dfrac{1}{v} + \dfrac{1}{-40} = \dfrac{1}{40}$

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

$\dfrac{1}{v} = \dfrac{1}{40} - \dfrac{1}{-40} =$

$\dfrac{1}{v} = \dfrac{2}{40}$

v = $\dfrac{40}{2}$

v = 20 cm

Now,

Magnification, m = $\dfrac{{h}_{i}}{{h}_{o}} = \dfrac{-v}{u}$

$\dfrac{{h}_{i}}{10} = \dfrac{-20}{-40}$

${h}_{i} = 5 cm$

Hence, The distance of the image from the pole is 20 cm.

Height of the image is 5 cm.

The image is virtual, erect and diminished.