Question

A concave mirror of focal length $15cm$ forms an image having twice the linear dimension of the object. The position of the object when the image is virtual will be
A. $22.5cm$
B. $7.5cm$
C. $30cm$
D. $45cm$

Answer 1

Hint-Here the magnification of the mirror is given as 2. We know that magnification is the ratio of image distance to object distance. Using this we can find the relationship between image distance and object distance. By substituting this value in the mirror formula, we can find the position of the object.

Complete step by step solution:
It is given that the focal length of a concave mirror is $15cm$ .
$\Rightarrow f=15cm$
It forms an image having twice the linear dimension of the object. We need to find the position of the object when the image formed is virtual.
Since it is given that the image formed is twice that of the original size, the magnification is two.
We know that magnification in a concave mirror can be found as the ratio of the image distance to the object distance.
$m=-{\displaystyle \frac{v}{u}}$
Where v is image distance and u is the object distance.
$\Rightarrow 2=-{\displaystyle \frac{v}{u}}$
$\Rightarrow v=-2u$
Now let us use the mirror formula.
Mirror formula is given as
${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{v}}$
Where, f is the focal length, u is the object distance and v is the image distance.

On substituting the value of focal length and image distance we get
$\Rightarrow {\displaystyle \frac{1}{-15}}={\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{-2u}}$
 A virtual image is the image formed behind the mirror. According to sign convention the distance from the pole of the mirror to the image which is behind the mirror is taken as positive. Distance in front of the mirror is taken as negative. So, the focal length which is in front of the mirror for a concave mirror should also have a negative sign.
On solving the equation, we get the value of u as
$\Rightarrow {\displaystyle \frac{1}{-15}}={\displaystyle \frac{1}{2u}}$
$\therefore u=-7.5cm$
The negative value shows that the object is placed in front of the mirror.

Therefore, the correct answer is option B.

Note:Remember that a virtual image is an image that is formed behind the mirror. It is called virtual because this image cannot be caught on a screen. Whereas an image formed in front of the mirror is called a real image. It can be caught on a screen. And according to the sign convention the distance from pole to an image which is behind the mirror is taken as positive. And the distance from pole to object in front of the mirror is taken as negative.