Radiation Pressure — Question 10

Section 33-5 · Physics pages 5470–5471

Question 10 A symmetric lossless deflector feels force despite zero heating A stationary optical device receives vacuum power P along +z and sends P/2 into each of two outgoing beams making angles plus and minus theta from +z in the xz plane. There is no absorption or backward beam. P P/2 P/2 +z Schematic; labels, not drawing lengths, specify the data. (a) Find the net force using vector momentum balance. (b) Evaluate the longitudinal force at theta=60 degrees and its small-angle approximation.
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Question 10 - Solution A symmetric lossless deflector feels force despite zero heating P P/2 P/2 +z (a) Method and calculation ˙pout = P 2c[(sin θˆx + cos θˆz) + (−sin θˆx + cos θˆz)]. F = P c (1 −cos θ)ˆz, Fx = 0 . (b) Interpretation and further calculation Fz(60◦) = P/(2c), ˙Q = 0 . 1 −cos θ ≃θ2/2 ⇒Fz ≃Pθ2/(2c) (θ ≪1). Physical check The output power equals the input power, but its longitudinal momentum is smaller; the device receives the difference.

Question 10

A symmetric lossless deflector feels force despite zero heating

A stationary optical device receives vacuum power P along +z and sends P/2 into each of two outgoing beams making angles plus and minus theta from +z in the xz plane. There is no absorption or backward beam.

Diagram shown in the original SVG worksheet.

  1. Find the net force using vector momentum balance.

  2. Evaluate the longitudinal force at theta=60 degrees and its small-angle approximation.

Question 10 - Solution

A symmetric lossless deflector feels force despite zero heating

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝒑̇out=P2c[(sin⁡θx̂+cos⁡θẑ)+(−sin⁡θx̂+cos⁡θẑ)].\dot{\mathbf p}_{\mathrm{out}}=\frac{P}{2c}[(\sin\theta\hat x+\cos\theta\hat z)+(-\sin\theta\hat x+\cos\theta\hat z)].𝑭=Pc(1−cos⁡θ)ẑ,Fx=0.\boxed{\mathbf F=\frac{P}{c}(1-\cos\theta)\hat z,\quad F_x=0}.

(b) Interpretation and further calculation

Fz(60∘)=P/(2c),Q̇=0.\boxed{F_z(60^\circ)=P/(2c),\quad\dot Q=0}.1−cos⁡θ≃θ2/2⇒Fz≃Pθ2/(2c)(θ≪1).1-\cos\theta\simeq\theta^2/2\Rightarrow\boxed{F_z\simeq P\theta^2/(2c)}\quad(\theta\ll1).

Physical check

The output power equals the input power, but its longitudinal momentum is smaller; the device receives the difference.