1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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
πR2Ic
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2πR2Ic
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2R4Ic(R2+H2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2πR4Ic(R2+H2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 2πR4Ic(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. Now, horizontal component of the force, dF′=dFsinθ ∴ Total Force F=∫dF′=∫dFsinθ =∫2Ic(dA′)sinθ×sinθ=2Icsin2θ∫dA′ ∴F=2Icsin2θ∫dA′ Where, ∫dA′=Projected area=AR2 ⇒F=2Icsin2θ×πR2 From figure, sinθ=R√R2+H2 ⇒F=2Ic×[R2R2+H2]×πR2 ∴F=2πIR4c(R2+H2) Hence, (D) is the correct answer.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Falling Balls in Disguise
PHYSICS
Watch in App
Join BYJU'S Learning Program