An object of length 2.0 cm is placed at a distance of 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. The length of the image will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -4.0 cm
Use the mirror formula with the Cartesian sign convention: distances measured against the incident light are negative, and heights above the principal axis are positive. For the concave mirror the focal length is \( -f \) and the object distance is \( u = -1.5f \).
\( \displaystyle \frac{1}{v} = \frac{1}{f_{\text{mirror}}} - \frac{1}{u} = -\frac{1}{f} + \frac{1}{1.5f} = -\frac{1}{3f} \)
\( \displaystyle v = -3f \)
The magnification is
\( \displaystyle m = -\frac{v}{u} = -\frac{-3f}{-1.5f} = -2 \)
\( \displaystyle h' = m\,h = -2 \times 2.0\,\mathrm{cm} = -4.0\,\mathrm{cm} \)
The object lies between the focus and the centre of curvature, so the image is real, inverted and twice as long; the minus sign shows that it is inverted.