Question

an object of height 5 cm is placed on the principal axis of a concave mirror of focal length 20cm is a virtual images formed by a concave mirror at a distance of 10 cm from its pole and find the size of image ....​

Answer 1

As we know that magnification=image distance(v)/object distance(u)
First of all focal length of a concave lens in always (-ve)
Thus f=-20cm
as given v=-10cm
By lens formula,
1/v-1/u=1/f
Substituting values into the equation we get
u=-20
Thus magnification is m=v/u=(-10)/(-20)=0.5
I’m sorry if I’m wrong

Answer 2

The height of the image is 2.5 cm.

Explanation:

Given that,

Height of the  object, h = 5 cm

Object distance, u = -20 cm

Image distance, v =10 cm (virtual image)

Let h' is the size of image. The magnification of mirror is given by :

$m=\dfrac{-v}{u}=\dfrac{h'}{h}$

$\dfrac{-(10)}{-20}=\dfrac{h'}{5}$

h' = +2.5 cm

So, the height of the image is 2.5 cm. Hence, this is the required solution.

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