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JEE Main 2021 · Physics · Ray and Wave Optics · hard

A plano-convex lens of refractive index 1.5 has radius of curvature 30 cm for its curved surface. Its plane face is silvered. A small object is…

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A plano-convex lens of refractive index 1.51.5 has radius of curvature 30 cm30\ \mathrm{cm} for its curved surface. Its plane face is silvered. A small object is placed 45 cm45\ \mathrm{cm} in front of the curved face on the axis. The final image is formed
  1. A. 90 cm90\ \mathrm{cm} in front of the curved face
  2. B. 18 cm18\ \mathrm{cm} in front of the curved face
  3. C. 30 cm30\ \mathrm{cm} in front of the curved face
  4. D. 90 cm90\ \mathrm{cm} behind the silvered face
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Answer: (A) 90 cm90\ \mathrm{cm} in front of the curved face

Light passes through the lens, reflects from the plane silvered face and passes through the lens again. Lens focal length: 1f=(1.5−1)130⇒f=60 cm\dfrac1f=(1.5-1)\dfrac1{30}\Rightarrow f=60\ \mathrm{cm}. The plane mirror contributes zero power, so the equivalent power is 1F=2f=130\dfrac1F=\dfrac2f=\dfrac1{30}: the system behaves as a concave mirror of focal length 30 cm30\ \mathrm{cm}. Mirror formula with u=−45u=-45, f=−30f=-30: 1v=−130+145=−190\dfrac1v=-\dfrac1{30}+\dfrac1{45}=-\dfrac1{90}, so v=−90 cmv=-90\ \mathrm{cm}, a real image 90 cm90\ \mathrm{cm} in front of the curved face.
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