A concave mirror has a focal length of 10 cm. An object can be placed at two different points in front of it so that the image is twice the size of the object. The distance between these two points is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10 cm
A concave mirror can give an image twice the size in two ways. The image may be real and inverted, with \( m=-2 \). It may also be virtual and erect, with \( m=+2 \). Each case gives one object position. Here \( f=-10\ \mathrm{cm} \).
For the real image, \( m=-2 \) gives \( v=2u \).
\( \begin{aligned} \frac{1}{2u}+\frac{1}{u} &= -\frac{1}{10} \\ \frac{3}{2u} &= -\frac{1}{10} \\ u &= -15\ \mathrm{cm} \end{aligned} \)
For the virtual image, \( m=+2 \) gives \( v=-2u \).
\( \begin{aligned} -\frac{1}{2u}+\frac{1}{u} &= -\frac{1}{10} \\ \frac{1}{2u} &= -\frac{1}{10} \\ u &= -5\ \mathrm{cm} \end{aligned} \)
\( \begin{aligned} d &= 15-5 \\ &= 10\ \mathrm{cm} \end{aligned} \)
Both points lie in front of the mirror, one between F and C and one between P and F. Hence, the two points are 10 cm apart.
NCERT (2026-27):
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.”
The mirror equation – “… so that”
\( \displaystyle m=\frac{h^{\prime}}{h}=-\frac{v}{u} \)\( \qquad (9.9) \)