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Diffraction and Interference Study Guide (page 2)

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Updated on Sep 27, 2011

Example

Consider two sources emitting waves and two waves meeting in a region of space. The wave diagram is shown in Figure 20.4. What is the result of the superposition of the two waves?

 

Superposition Principle

Solution

At each point in the figure, the two displacements are equal and opposite. Because the superposition is an algebraic sum of the two waves, the result will be zero at every point, as shown by the example on the left:

A + (–A) = 0

y1 = 20 cm · sin (1.1 · 1010 · x – 2 · t)

y2 = 23 cm · sin (0.94 · 1010 · x – 0.18 · t)

Diffraction and Huygens' Principle

If the lights are turned on in your house and you close your bedroom door, you can still tell the lights are on. How does this happen? If light only propagates in a straight line, then there should be only a small amount making its way through the keyhole or under the door. Is there something more to propagation?

The answer is diffraction, which is the bending of waves around obstacles. Bending occurs proportionally to the wavelength of the wave and is inverse proportional to the size of the obstacle. That means that for light waves (which have small wavelengths) and large openings, the ratio of the wavelength to the width (λ/w) will be very small, and the bending will not permit you to see around a corner. If we imagine the same source of waves producing some plane waves that encounter a small and a large obstacle, beyond the obstacle, diffraction will create different patterns of wave propagation, as shown in Figure 20.6. For a largesize aperture (w > > λ), no bending is produced, whereas when the two are comparable (w ~ λ), bending occurs.

Diffraction and Huygens' Principle

The waves we see on the right of Figure 20.6 are actually composed by superimposing many waves produced by different points at the obstacle. Huygens' Principle states that every point on a wavefront produces a subset of waves that moves with the same velocity as the incident wave. At a later time, the wavefront is found to be the surface tangent to the propagating subset of waves (see Figure 20.7).

Diffraction and Huygens' Principle

Example 1

Will red light or violet light produce a broader fringe pattern on a screen? (See Figure 20.7.)

Solution 1

The size of the broadening pattern will be larger when more diffraction will take place. The diffraction pattern is dependent on the wavelength and the width of the opening:

Fringe pattern ~

At the same size of aperture (opening), the red light, which has a larger wavelength, will give a larger diffraction pattern, while violet (at the other end of the spectrum), with a reduced wavelength, will show a less extended fringe pattern.

Figure 20.7 not only portrays Huygens' Principle; it also shows the image formed on a screen: a strong central light followed by symmetric fringes of dark and light. Considering the earlier example with the waves that cancel each other, it will make perfect sense to say that at some points on the screen, the two or more waves present at the same time will be in opposite phase. Hence, when superposition adds the waves, the result will be zero: no light. At other positions, the resultant will be a nonzero amount and even reach a maximum when the superposition is among waves that are in phase with each other.

We call the situation when a dark fringe (no light) is obtained destructive interference, and constructive interference leads to light. The condition for destructive interference is

where m equals 1, 2, 3, and so on depending on which fringe the angle θ is considered for: the first dark fringe from the bright central spot, the second, third, and so on (see Figure 20.8).

Diffraction and Huygens' Principle

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