Question

Find the nature and position of the image of an object placed 10 cm from a diverging
lens of focal length 15 cm?

Answer 1

(3+√3) (3-√3)(3+√3) (3-√3)(3+√3) (3-√3)

Answer 2

Answer:

v=6cm

Explanation:

It is to be remembered that a convex mirror always forms the virtual, erect and diminished image.

Given: Object distance, u=−10cm     Focal length, f=15cm

To find: Image distance, v

From mirror formula :

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

$\frac{1}{v}$ = $\frac{1}{15}$ + $\frac{1}{10}$ = $\frac{25}{100}$

v=6cm

So,  image will be formed 6 cm beyond the mirror (virtual magnification)

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

So the image be erect and diminished.

HOPE IT HELPS!!