An object O is placed at a distance of 0.2 m to the left of a convex lens of focal length 0.1 m. A concave mirror of focal length 0.05 m is placed at a distance of 0.3 m to the right of the lens. The nature and position of the final image will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → real and inverted at the position of the object O itself
Trace the light through the lens, the mirror and the lens again (Cartesian sign convention). Refraction at the lens (u = −0.2 m, f = +0.1 m):
\( \displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{0.1} - \frac{1}{0.2} = \frac{1}{0.2} \)
So \( v = +0.2\,\mathrm{m} \): a real, inverted image at 2F, to the right of the lens.
Reflection at the mirror: this image is \( 0.3 - 0.2 = 0.1\,\mathrm{m} \) in front of it, and \( R = 2f = 0.1\,\mathrm{m} \), so it lies at the centre of curvature; the mirror forms a real, inverted image at the same place.
Refraction at the lens again: this image is again at 2F, so the lens forms a real image at 2F on its left, at the position of O.
Three inversions leave the final image inverted, real and the same size as O.