Two light sources can be adjusted to emit monochromatic...

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, $2.04 \mu \mathrm{m}$ apart, and in line with an observer, so that one source is $2.04 \mu \mathrm{m}$ farther from the observer than the other. (a) For what visible wavelengths $(380$ to $750 \mathrm{nm})$ will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is $2.04 \mu \mathrm{m}$ farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

Key Concepts

Observation Geometry

The geometry of the source and observer arrangement influences the determination of the path difference. Whether the sources are positioned directly in line with the observer or arranged at an angle, the essential factor remains the extra distance one wave travels relative to the other. If the same path difference is maintained, the conditions for constructive and destructive interference remain unchanged.

Constructive Interference

Constructive interference occurs when the waves from the two light sources arrive in phase at the observer, which happens when the path difference is an integer multiple of the light wavelength (?d = m?, where m is an integer). This phase alignment results in the enhancement of the wave amplitudes, leading to a brighter, more intense light at that specific wavelength and location.

Destructive Interference

Destructive interference takes place when the waves arrive out of phase, specifically when the path difference is an odd multiple of one-half the wavelength (?d = (m + ½)?, with m being an integer). In this scenario, the wave crests and troughs cancel each other out, resulting in a significant reduction or complete cancellation of the light intensity at the point of observation.

Monochromatic Light in the Visible Spectrum

Monochromatic light consists of waves that have only a single wavelength, which in the context of the visible spectrum ranges from about 380 to 750 nm. This narrow spectral emission allows for clear analysis of interference effects, such as the exact conditions required for bright (constructive) or dark (destructive) fringes, without complications from multiple wavelengths.

Coherent Light Sources

Coherent light sources have a fixed phase relationship, meaning that the waves they produce maintain constant phase differences. This quality is essential for creating stable interference patterns, whether constructive or destructive, as only coherent sources can produce a consistent relationship between the wavefronts over time.

Path Difference

The path difference refers to the difference in the distance traveled by light from two sources to the observer. It is a critical parameter in determining whether the waves will interfere constructively or destructively. A given path difference leads to a specific phase difference between the waves when they meet, thereby controlling the interference outcome.

Transcript

00:01 This is chapter 35, problem number six.

00:03 We have two light sources, a and b.

00:06 And then we have the observer.

00:08 The distance between a and b is 2 .04 micrometers.

00:13 And the distance from the other source to the observer is 2 .04 micrometers again.

00:22 Now, in part a, we're asked to calculate in what visible vague length, we would have constructive interference for this sort of alignment.

00:34 And as you know, if you want constructive interference, the difference in path has to be, path difference has to be equal to m times an integer times lambda.

00:55 So in this case, we're the observer.

00:57 So r a, let's say, and this is rb, the difference in paths would be r a minus r b that equals n times lambda so r a is actually 4 .08 micrometers minus r b is the half of it right so then we have 2 .04 micrometers distance a path difference and that if we write it in nanometers it's going to be in the um near inferior region.

01:35 So this has to be equal to n times lambda, but we're looking for the bay length that fall into the visible range.

01:45 So for that, for example, if we take m equals one, our lambda from this equation is going to be somewhere in the near inferior region, but we don't want that.

01:57 So not invisible, right? not in the visible range.

02:00 We want the emission in the visible range.

02:04 So then if you do m equals 2, you're still going to see that it's 10, 20 nanometers, still in the near infrared.

02:11 So if we go to m equals 3, though, what we have is 3 times lambda equals 240 nanometers.

02:20 So then from lambda here is going to be 680 nanometers, which is in the visible range, right? so this is one of the answers...

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