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NVAMediumJEE 2023Photoelectric Effect

JEE Physics 2023 Question with Solution

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24 W24 \, \text{W}. The radius of curvature of the hemisphere is 10 cm10 \, \text{cm} and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is _____ × 10−810^{-8} N.

Answer

Correct answer:4

Step-by-step solution

the solution unavailable

Given: Power emitted by the source is 24 W24 \, \text{W}. Radius of the hemispherical reflecting surface is 10 cm10 \, \text{cm}.

Find: Force on the hemisphere due to incident light pressure.

The solution is unrelated to this question, so the working could not be extracted from the solution. The answer has therefore been taken from the provided correct answer field as 44.

Therefore, the required numerical value is 44.

Common mistakes

  • Using the full spherical area instead of the hemispherical reflecting surface is incorrect because only the hemisphere receives and reflects the light. Use the geometry of the hemisphere actually present in the question.

  • Treating the surface as absorbing instead of completely reflecting is wrong because momentum change doubles on reflection. Use the reflecting-surface relation for radiation pressure or force.

  • Ignoring the direction of momentum transfer over the curved surface leads to an incorrect net force. Resolve the contribution using symmetry so that only the axial components add.

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