An object placed in front of a concave mirror of focal length \(20\) cm makes (i) a real image and (ii) a virtual image with magnification \(4\) in both the cases. The position of the object in two cases respectively, will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-25\,\mathrm{cm}, -15\,\mathrm{cm}\)
For a mirror, with the Cartesian sign convention (distances measured from the pole, positive in the direction of the incident light), \( f = -20\,\mathrm{cm} \) for the concave mirror and
\( \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f}, \qquad m = -\frac{v}{u} \)
A real image is inverted \( (m = -4) \); a virtual image is erect \( (m = +4) \).
(i) Real image, \( m = -4 \), so \( v = 4u \):
\( \displaystyle \frac{1}{4u} + \frac{1}{u} = \frac{5}{4u} = -\frac{1}{20} \)
\( \displaystyle u = -25\,\mathrm{cm} \)
(ii) Virtual image, \( m = +4 \), so \( v = -4u \):
\( \displaystyle -\frac{1}{4u} + \frac{1}{u} = \frac{3}{4u} = -\frac{1}{20} \)
\( \displaystyle u = -15\,\mathrm{cm} \)
The object is beyond F for the real image and between F and the pole for the virtual image, as expected.