Radiation Pressure — Question 5

Section 33-5 · Physics pages 5460–5461

Question 5 Tilting a reflecting sheet changes the vertical support force cubically A perfectly reflecting sheet has area A and mass per area sigma. A wide beam travels vertically upward with intensity I. The sheet normal makes angle theta with vertical. A lateral constraint prevents sideways motion without providing vertical force. I ˆn Mg Schematic; labels, not drawing lengths, specify the data. (a) Derive the vertical optical force and intensity required to balance weight. (b) Find the intensity multiplier when the sheet is tilted from zero to 60 degrees.
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Question 5 - Solution Tilting a reflecting sheet changes the vertical support force cubically I ˆn Mg (a) Method and calculation Phit = IA cos θ, Fn = 2IA cos2 θ/c. Fz = Fn cos θ = 2IA cos3 θ/c. (b) Interpretation and further calculation Fz = σAg ⇒Ibalance = σgc 2 cos3 θ . I(60◦)/I(0) = sec3 60◦= 8 . Physical check This is a force-balance condition, not a claim of stable levitation; the horizontal reaction is supplied by the stated constraint.

Question 5

Tilting a reflecting sheet changes the vertical support force cubically

A perfectly reflecting sheet has area A and mass per area sigma. A wide beam travels vertically upward with intensity I. The sheet normal makes angle theta with vertical. A lateral constraint prevents sideways motion without providing vertical force.

Diagram shown in the original SVG worksheet.

  1. Derive the vertical optical force and intensity required to balance weight.

  2. Find the intensity multiplier when the sheet is tilted from zero to 60 degrees.

Question 5 - Solution

Tilting a reflecting sheet changes the vertical support force cubically

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Phit=IAcos⁡θ,Fn=2IAcos⁡2θ/c.P_{\mathrm{hit}}=IA\cos\theta,\quad F_n=2IA\cos^2\theta/c.Fz=Fncos⁡θ=2IAcos⁡3θ/c.F_z=F_n\cos\theta=2IA\cos^3\theta/c.

(b) Interpretation and further calculation

Fz=σAg⇒Ibalance=σgc2cos⁡3θ.F_z=\sigma Ag\Rightarrow\boxed{I_{\mathrm{balance}}=\frac{\sigma gc}{2\cos^3\theta}}.I(60∘)/I(0)=sec⁡360∘=8.\boxed{I(60^\circ)/I(0)=\sec^3 60^\circ=8}.

Physical check

This is a force-balance condition, not a claim of stable levitation; the horizontal reaction is supplied by the stated constraint.