Intensity in Double-Slit Interference — Question 9

Section 35-5 · Physics pages 5808–5809

Question 9 Complementary output ports make interference energy accounting explicit An ideal lossless balanced interferometer has output powers P+ = Pin(1 + cos δ)/2 and P−= Pin(1 −cos δ)/2. Its input power is 12 mW. Assume no losses and identical detector calibration. Pin P+ P− δ Schematic; labels, not drawing lengths, specify the data. (a) Calculate both powers at δ = π/3 and verify total-power conservation. (b) Derive the normalized difference signal and find both phase possibilities in [0, 2π) if the measured normalized difference is −1/2.
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Question 9 - Solution Complementary output ports make interference energy accounting explicit Pin P+ P− δ (a) Method and calculation P+ = 9.0 mW, P−= 3.0 mW . P+ + P−= Pin =⇒Ptotal = 12 mW . (b) Interpretation and further calculation D = P+ −P− P+ + P− = cos δ . D = −1/2 =⇒δ = 2π/3, 4π/3 . Physical check Destructive interference at one port is accompanied by increased power at the other port; it does not destroy electromagnetic energy.

Question 9

Complementary output ports make interference energy accounting explicit

An ideal lossless balanced interferometer has output powers P+=Pin(1+cos⁡δ)/2P_+=P_{\mathrm{in}}(1+\cos\delta)/2 and P−=Pin(1−cos⁡δ)/2P_-=P_{\mathrm{in}}(1-\cos\delta)/2. Its input power is 12mW12\,\mathrm{mW}. Assume no losses and identical detector calibration.

Diagram shown in the original SVG worksheet.

  1. Calculate both powers at δ=π/3\delta=\pi/3 and verify total-power conservation.

  2. Derive the normalized difference signal and find both phase possibilities in [0,2π)[0,2\pi) if the measured normalized difference is −1/2-1/2.

Question 9 - Solution

Complementary output ports make interference energy accounting explicit

Diagram shown in the original SVG worksheet.

(a) Method and calculation

P+=9.0mW,P−=3.0mW.\boxed{P_+=9.0\,\mathrm{mW},\quad P_-=3.0\,\mathrm{mW}}.P++P−=Pin⇒Ptotal=12mW.P_++P_-=P_{\mathrm{in}}\Longrightarrow\boxed{P_{\mathrm{total}}=12\,\mathrm{mW}}.

(b) Interpretation and further calculation

D=P+−P−P++P−=cos⁡δ.\boxed{D=\frac{P_+-P_-}{P_++P_-}=\cos\delta}.D=−1/2⇒δ=2π/3,4π/3.D=-1/2\Longrightarrow\boxed{\delta=2\pi/3,\ 4\pi/3}.

Physical check

Destructive interference at one port is accompanied by increased power at the other port; it does not destroy electromagnetic energy.