A concave mirror of focal length 30 cm forms a real image of an object on a screen. The screen is 40 cm from the mirror. The magnification of the image is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\frac{1}{3}\)
The real image falls on the screen in front of the mirror, so \( v=-40\ \mathrm{cm} \). The mirror equation gives the object distance. Then \( m=-v/u \) gives the magnification.
\( f=-30\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{u} &= \frac{1}{f}-\frac{1}{v} \\ &= -\frac{1}{30}+\frac{1}{40} \\ &= -\frac{1}{120} \end{aligned} \)
\( u=-120\ \mathrm{cm} \)
\( \begin{aligned} m &= -\frac{v}{u} \\ &= -\frac{(-40)}{(-120)} \\ &= -\frac{1}{3} \end{aligned} \)
The negative sign shows an inverted image. Hence, the image is one-third the size of the object.
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”
\( \displaystyle m=\frac{h^{\prime}}{h}=-\frac{v}{u} \)\( \qquad (9.9) \)