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Q.
An object is placed at 15 cm in front of a concave mirror whose focal length is $10 cm$. The image formed will be
Ray Optics and Optical Instruments
Solution:
According to Cartesian sign convention
Object distance, $u=-15 \,cm$, Focal length, $f=-10 \,cm$
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 $v=-30\, cm$
The image is $30 \,cm$ from the mirror on the same side of the object.
Magnification, $m=-\frac{v}{u}=-\frac{(-30\, cm )}{(-15 \,cm )}=-2$
The image is magnified, real and inverted.