An object is placed 10 cm from a lens and a real image is formed with a magnification of magnitude 0.5. The lens is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → convex, with focal length \( \frac{10}{3}\,\mathrm{cm} \)
A real image formed by a single lens is always inverted. So, with the Cartesian sign convention, the magnification \( m=v/u \) is negative, that is, \( m=-0.5 \). The focal length then follows from the lens formula \( 1/v-1/u=1/f \).
The object is on the left of the lens:
\( u=-10\,\mathrm{cm} \)
The image distance is:
\( \begin{aligned} v &= mu \\ &= (-0.5)\times(-10\,\mathrm{cm}) \\ &= +5\,\mathrm{cm} \end{aligned} \)
The lens formula gives:
\( \begin{aligned} \frac{1}{f} &= \frac{1}{v}-\frac{1}{u} \\ &= \frac{1}{+5\,\mathrm{cm}}-\frac{1}{-10\,\mathrm{cm}} \\ &= \frac{2+1}{10\,\mathrm{cm}} \\ &= \frac{3}{10\,\mathrm{cm}} \end{aligned} \)
\( \displaystyle f=\frac{10}{3}\,\mathrm{cm} \)
The focal length is positive, so the lens is a convex (converging) lens.