Question

find the position of an object when placed in front of a concave mirror of focal length 20 cm produces a virtual image,which is twice size of object.

Answer 1

Answer:

The image is at distance of 20 cm from the mirror on the side opposite to that of object.

Explanation:

Given,

f = −20 and M = 2

We know that ;

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

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

=$v = \frac{uf}{u-f}$

We know,

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

Substituting the value of v ,

M = - $\frac{f}{u-f}$

Substituting value of f = 20

M = $\frac{20}{u+20}$

According to question,

M = 2

$\frac{20}{u+20}$ = 2

⇒20 = 2u + 40

⇒ u = − 10.

⇒$v = \frac{uf}{u-f}$

⇒$v = \frac{10*20}{-10+20}$  = 20

Hence,

The image is at distance of 20 cm from the mirror on the side opposite to that of object.

Answer 2

Hello dear

Explanation:

Here is the answer---

Given ⇒

Focal Length of the Mirror = 20 cm. (negative)

Let the Height of the Object be H.

∴ Height of the Image = 2H

Now, 

Magnification = H₁/H₀

 = 2H/H

 = 2

Also, m = -v/u

2 = -v/u

⇒  v = -2u

Now,

Image Distance = -2u 

Object distance = u

 

Using the Mirror Formula,

 

=>1/f = 1/v+1/u 

=> 1/-20= 1/-(-2u)+1/u

=> 1/-20 = 1+2/2u

=> 2u = -20 × 3

=> 2u = -60

=> u = -30 cm.

∴ Distance of the Object from the mirror is 30 cm. Negative size shows that the object is kept on the side of the mirror.

Hope it helps !!!