Question

An object is placed at 15 cm in front of a concave mirror whose focal length is $10cm$. The image formed will be

• A. magnified and inverted
• B. magnified and erect
• C. reduced in size and inverted
• D. reduced in size and erect.

Answer 1

Answer: (A)

According to Cartesian sign convention
Object distance, $u=-15cm$, Focal length, $f=-10cm$
Using mirror formula, $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\Rightarrow \frac{1}{(-15)}+\frac{1}{v}=\frac{1}{(-10)}$
$\frac{1}{v}=\frac{1}{(-10)}-\frac{1}{(-15)}=\frac{1}{(-10)}+\frac{1}{(15)}$ or $v=-30cm$
The image is $30cm$ from the mirror on the same side of the object.
Magnification, $m=-\frac{v}{u}=-\frac{(-30cm)}{(-15cm)}=-2$
The image is magnified, real and inverted.