An object is placed at 5 cm in front of a concave mirror with a radius of curvature of 20 cm. The distance of the image from mirror and its nature are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 10 cm, virtual and erect
Use the mirror formula with the Cartesian sign convention (distances measured from the pole, positive in the direction of the incident light, which travels from the object towards the mirror).
\( f = -R/2 = -20/2 = -10\,\mathrm{cm} \)
\( u = -5\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-5} = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10} \)
\( \displaystyle v = +10\,\mathrm{cm} \)
A positive \( v \) means the image is 10 cm behind the mirror, so it is virtual.
\( \displaystyle m = -\frac{v}{u} = -\frac{10}{-5} = +2 \)
The positive magnification means the image is erect (and twice the size). This is expected: the object lies between the pole and the focus.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
The Mirror Equation – …
