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JEE Main 2024 · Physics · Ray and Wave Optics · hard

A convex lens of focal length 20 cm has a plane mirror placed 15 cm behind it, perpendicular to the principal axis. A point object is placed 30 cm in…

Single correct option

A convex lens of focal length 20 cm20\ \mathrm{cm} has a plane mirror placed 15 cm15\ \mathrm{cm} behind it, perpendicular to the principal axis. A point object is placed 30 cm30\ \mathrm{cm} in front of the lens on the axis. After refraction through the lens, reflection from the mirror and refraction again through the lens, the final image forms
  1. A. 60 cm60\ \mathrm{cm} from the lens, on the mirror side
  2. B. 12 cm12\ \mathrm{cm} from the lens, on the object side
  3. C. 30 cm30\ \mathrm{cm} from the lens, on the object side
  4. D. 12 cm12\ \mathrm{cm} from the lens, on the mirror side
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Answer: (B) 12 cm12\ \mathrm{cm} from the lens, on the object side

First refraction: 1v=120−130\dfrac1v=\dfrac1{20}-\dfrac1{30}, so v=60 cmv=60\ \mathrm{cm} behind the lens, i.e. 45 cm45\ \mathrm{cm} beyond the mirror at 15 cm15\ \mathrm{cm}. The mirror treats this as a virtual object and forms a real image 45 cm45\ \mathrm{cm} in front of it, at 15−45=−3015-45=-30, i.e. 30 cm30\ \mathrm{cm} to the left of the lens. Light now travels leftwards and is converging towards this point: for the second pass the object is virtual, u=+30 cmu=+30\ \mathrm{cm}, so 1v=120+130=112\dfrac1v=\dfrac1{20}+\dfrac1{30}=\dfrac1{12}, v=12 cmv=12\ \mathrm{cm} on the object side.
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