Question

A 3 cm object is placed in front of a concave mirror and the image formed is 9 cm in height. What is the magnification of this object? In general terms where was the object placed (beyond C, at C, between C and F, or between F and the mirror)?

Answer 1

Given:

The height of the object(O) = 3 cm.

The height of the image (I) = 9 cm.

To find:

• The magnificent of the image.
• The position of the object.

Solution:

The height of the object(O) = 3 cm.

The height of the image (I) = 9 cm.

Magnificent (m) = ?

From the formula we get that,

$m = \dfrac{I}{O} \\\\= \dfrac{9}{3}\\ \\= 3$

[ Since, magnification is a ratio so, it has no unit.]

∴ The magnificent of the image is 3.

Since,the magnificent of the image is positive hence, it is a virtual image. As it is a virtual image so, the object is placed on the focal length of the mirror (F).

So, the object is placed at F.

Answer:

           The magnificent of the image is 3.

            The object is placed at F.