Question

Hi

Answer 1

Explanation:

m = h₂/h₁

Here,

• m = Magnification
• h₁ = Height of the object
• h₂ = Height of the image formed by the concave mirror

Magnification given is (-3). The linear magnification here is negative.

This implies that h₂ < h₁ because m < 1.

Therefore, the image formed is real and inverted. It is smaller than the object. So, the image is formed between the Centre of Curvature & the Focus.

[NOTE: If the image is highly reduced, it is formed at the focus.]

Answer 2

Answer:HERE IS YOUR ANSWER

THE POINTS ARE

IF WE PUT THE OBJECT BETWEEN THE FOCUS

AND THE POLE

THEN THE IMAGE FORMED WILL be MAGNIFIED

If that’s not what you are looking for, try this one:

For concave mirror the virtual image is formed when the object is kept in between the pole and the focus.  

Given here the "size of the image" is twice to that of the object.

Hence, it is consider that the magnification is +2.

So, the magnification value is positive and the image formed will be "virtual and erect". Thus, the object should be kept in between the "pole and the focus" in concave mirror."

Explanation: