Interference and Energy Conservation

When two equal-amplitude waves arrive exactly out of phase ([\phi=(2n+1)\pi]) their electric fields cancel point-by-point: [ E = E_0\sin\omega t + E_0\sin(\omega t+\phi)=0.] Since intensity [I\propto E^2], the instantaneous field is zero ⇒ intensity is zero at that location. The waves themselves continue to exist elsewhere; energy is redistributed to bright fringes.

Energy conservation holds overall. Interference redistributes energy spatially: bright fringes have higher intensity than each source alone, while dark fringes have lower (or zero) intensity. Averaging intensity over many fringes equals the sum of intensities of the separate sources. For two equal sources [I_1=I_2=I_0], spatial average [ \bar I = I_1+I_2 = 2I_0].

Phase difference [\Delta\phi] directly relates to path difference [\Delta x] by [\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x]. Phase encodes relative oscillation at the instant of superposition, which determines whether fields add constructively or destructively — it is the natural quantity for interference.

Complete cancellation requires equal amplitudes and opposite phase. If amplitudes differ, resultant minimum is nonzero. Using vector addition: [ I_\text{min} \propto (A_1-A_2)^2,] so only equal amplitudes [A_1=A_2] can give zero.

No. Photons are a quantum description; destructive interference means probability amplitudes cancel at that point so detection probability is zero there. Energy is redirected — quantum mechanically, detection probabilities are redistributed, not destroyed.

Instantaneous intensity for two equal waves with phase [\phi] is [I=4I_0\cos^2(\tfrac{\phi}{2})] locally. Averaging over a whole interference cycle or many fringes (or phases) gives: [ I = 4I_0\cos^2(\tfrac{\phi}{2})] [= 2I_0 = I_1+I_2.] So total (time- or space-averaged) energy equals input energy.

No. The sources deliver the same average power. Interference rearranges where power is delivered (space distribution), but the net power radiated (integrated over all space) remains fixed by the sources.

Coherence (stable phase relation) is required to form steady interference patterns. If sources are incoherent, phases vary randomly and interference averages out; energy is simply the sum of intensities everywhere — conservation still holds, but no redistribution into fixed fringes occurs.

For two equal waves of amplitude [E_0]: maxima occur at [\phi=0] giving resultant field [2E_0] so intensity [I\propto(2E_0)^2 = 4E_0^2]. If an individual source intensity is [I_0\propto E_0^2], the bright fringe reaches [4I_0].

Yes: because energy is redistributed. For stationary coherent sources the interference pattern consists of alternating constructive and destructive regions; destructive spots correspond to constructive regions elsewhere so total energy is conserved.

A glass plate changes optical path and thus local phase. That shifts fringe positions: dark regions move and bright regions appear where dark used to be. Energy flows into newly formed bright regions; total energy remains constant.

Not permanently in free space from continuous sources — dark regions are stationary if coherence is maintained, but energy is not lost; it is absent only at those spatial points and appears elsewhere.

Yes. At interfaces some energy may be reflected/absorbed, reducing transmitted power. Interference redistributes what is transmitted; absorption reduces total transmitted energy, but overall conservation (including absorbed + reflected + transmitted) still holds.

Only components with the same polarization interfere. Orthogonally polarized waves do not produce interference fringes; energy simply adds. So effective redistribution requires overlapping polarization components.

Redistribution occurs at the speed of wave propagation — the pattern establishes after waves have had time to reach the observation region and for steady phase relations to form. It is not a global instantaneous transfer of energy; local fields adjust causally.

No. Dark fringes are locations of destructive interference where the local intensity is low or zero; energy is redistributed into bright fringes. Integrating intensity over the whole pattern returns the input energy.

Apparent local increase is compensated by decreases elsewhere. The **total** energy (integral of intensity) is conserved.

No. Different amplitudes still interfere; fringe visibility decreases but redistribution still occurs according to vector addition of fields.

False. Incoherent sources have rapidly varying phase; any instantaneous interference averages out and no stable pattern forms.

Locally instantaneous field may be zero, but there is no net “hole” in transmitted energy — energy flows around and concentrates elsewhere; detectors placed at different locations will register redistributed energy.

For stable, long-lived fringes ideally yes (strict coherence). Small frequency differences produce beats or moving fringes; averaging washes out the pattern.

Quantum interference redistributes detection probabilities but does not violate conservation laws — total expected detections equal source emission rates.

A screen simply samples the redistributed energy. Placing absorbers changes local energy (absorbed), but redistribution among transmitted regions still governed by interference prior to absorption.

No. Constructive regions are brighter because destructive regions are darker; the net energy remains the same.

No — interference is a general wave phenomenon (sound, water, electromagnetic, matter waves).