An object is placed at a distance of 10 cm from a concave mirror of focal length 15 cm. The magnification of the image is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3 \)
Use the mirror formula with the Cartesian sign convention (distances measured from the pole, positive along the incident light). For a concave mirror in front of which the object is placed:
\( u = -10\,\mathrm{cm} \)
\( f = -15\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{15} + \frac{1}{10} = \frac{1}{30} \)
\( \displaystyle v = +30\,\mathrm{cm} \)
\( \displaystyle m = -\frac{v}{u} = -\frac{30}{-10} = +3 \)
The object is inside the focus, so the image is virtual, erect (positive \( m \)) and three times magnified.
Shortcut: The relation \( m = f/(f - u) \) gives the magnification without finding \( v \):
\( \displaystyle m = \frac{-15}{-15 - (-10)} = \frac{-15}{-5} = 3 \)