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JEE Main 2020 · Physics · Ray and Wave Optics · hard

A solid glass sphere of refractive index 1.5 and radius 10 cm has a point source on its diameter, outside the sphere, 30 cm from the nearest surface.…

Single correct option

A solid glass sphere of refractive index 1.51.5 and radius 10 cm10\ \mathrm{cm} has a point source on its diameter, outside the sphere, 30 cm30\ \mathrm{cm} from the nearest surface. Considering paraxial rays only, the final image is formed at a distance from the farther surface, on the far side, equal to
  1. A. 9 cm9\ \mathrm{cm}
  2. B. 20 cm20\ \mathrm{cm}
  3. C. 70 cm70\ \mathrm{cm}
  4. D. 14 cm14\ \mathrm{cm}
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Answer: (D) 14 cm14\ \mathrm{cm}

First surface (air to glass): 1.5v−1−30=1.5−110\dfrac{1.5}{v} - \dfrac{1}{-30} = \dfrac{1.5-1}{10}, so 1.5v=0.05−0.0333\dfrac{1.5}{v} = 0.05 - 0.0333 and v=90 cmv = 90\ \mathrm{cm} from the first surface. The second surface is 20 cm20\ \mathrm{cm} away, so it receives converging light directed to a point 70 cm70\ \mathrm{cm} beyond it, a virtual object. Second surface (glass to air, radius −10 cm-10\ \mathrm{cm}): 1v′−1.570=1−1.5−10=0.05\dfrac{1}{v'} - \dfrac{1.5}{70} = \dfrac{1-1.5}{-10} = 0.05, so 1v′=0.0714\dfrac1{v'} = 0.0714 and v′=14 cmv' = 14\ \mathrm{cm} beyond the farther surface. The value 7070 is the intermediate virtual object distance.
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