Radiation force experienced by a body (shown in the figure) exposed to radiation of intensity I assuming the surface of the body to be perfectly reflecting is (c is the velocity of light)
A
2πR4Ic(R2+H2)
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B
2R4Ic(R2+H2)
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C
πR2Ic
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D
2πR2Ic
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Solution
The correct option is A2πR4Ic(R2+H2)
The change in momentum of photons incident on the elemental area dA, along the perpendicular direction of the reflecting surface, is
ΔP=2psinθ=2(hλ)sinθ...(1)
Total number of photon incident per unit of time,
dn=I(dA′)E=I(dA′)(hcλ)
Force on the element :
∴dF=dh(Δp)
=I(dA′)(hcλ)×(2hλ)sinθ=2Ic(dA′)sinθ
Since, the vertical components of dF over the elements cancel each other due to symmetry shape.