An object is placed 5 cm in front of a concave mirror of radius of curvature 15 cm. The distance of the image from the mirror is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 15 cm
The focal length of a spherical mirror is half its radius of curvature. Distances measured in front of a concave mirror are taken as negative. The image position then follows from the mirror equation.
Given:
\( u=-5\ \mathrm{cm} \)
\( R=-15\ \mathrm{cm} \)
\( \begin{aligned} f &= \frac{R}{2} \\ &= -7.5\ \mathrm{cm} \end{aligned} \)
From the mirror equation:
\( \begin{aligned} \frac{1}{v} &= \frac{1}{f}-\frac{1}{u} \\ &= -\frac{1}{7.5}+\frac{1}{5} \\ &= \frac{-2+3}{15} \\ &= \frac{1}{15}\ \mathrm{cm^{-1}} \end{aligned} \)
\( v=+15\ \mathrm{cm} \)
From the above, \( v \) is positive, so the image is 15 cm behind the mirror. The object lies between the pole and the focus, so the image is virtual and erect. Hence, the image is 15 cm from the mirror.
NCERT (2026-27):
Focal length of spherical mirrors – “We now show that f = R/2, where R is the radius of curvature of the mirror. …”
The mirror equation – “Dividing it by v, we get”
\( \displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \qquad (9.7) \)
“This relation is known as the mirror equation.”