Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become

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Q: 80 (NDA-II/2020)
Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,2,2,2,2,0,0

When light of uniform intensity impinges perpendicularly on a totally reflecting surface, the radiation force on the surface is twice the momentum transfer per second.

The radiation force is given by the equation F = 2I/c, where F is the force, I is the intensity of the light, and c is the speed of light.

If the area of the surface is halved, the intensity of the light impinging on the surface will also be halved. This is because intensity is defined as power per unit area, so if the area is halved, the power remains the same but is spread over a smaller area.

Substituting the new intensity into the equation for the radiation force, we have F = 2(I/2)/c = I/c.

Thus, the force will be halved when the area is halved.

Therefore, the correct answer is option 2: half.

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