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Question

If a photon of intensity I falls on a surface at an angle 60 making with it having absorption coefficient 0.4, then radiation pressure exerted on the surface is-

A
0.4Ic
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B
1.2Ic
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C
1.05Ic
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D
7I20c
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Solution

The correct option is B 1.2Ic

When a parallel beam of radiation falls on a plane surface at an angle θ with normal to the surface, if the surface partially reflects (at same angle) and absorbs, then force exerted on surface is,

F=IAc(1+r)cosθ

Component of force normal to surface is,

F=Fcosθ=IAc(1+r)cos2θ

Radiation pressure = FcosθA=I(1+r)cos2θc

Given:
α=0.4r=1α=0.6
ϕ=60θ=90ϕ=30

P=Ic(1+0.6)×(32)2=1.2Ic

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (D) is the correct answer.
Why this question?

Caution: Most of the time, student repeat the mistake while putting the value of angle θ. It is the angle between the incident or reflected ray and normal to the surface.

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