Question

An object placed in front of a concave mirror of focal length $0.15m$ produces a virtual image, which is twice the size of the object. The position of the object with respect to the mirror is :

• A. $-5.5cm$
• B. $-6.5cm$
• C. $-7.5cm$
• D. $-8.5cm$

Answer 1

The correct option is C $-7.5cm$

Given:

Focal length $f=-0.15$m (concave mirror)

Magnification $m=2$

Let the position of the object is $=-u$

Since the image is virtual, it is erect.

So, magnification $m=-\frac{v}{u}=-\frac{v}{(-u)}=2$

$\Rightarrow v=2u$

Now, by mirror formulae

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

$\Rightarrow \frac{1}{2u}+\frac{1}{-u}=\frac{1}{-0.15}$

$\Rightarrow -\frac{1}{2u}=\frac{1}{-0.15}$

$\Rightarrow u=\frac{0.15}{2}$m

$\Rightarrow u=\frac{15}{2}$cm$=7.5$cm

Sign convention has been already used, So $u=-7.5cm$

So, Ans is (C).