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Chapter 28

Physical Optics: Interference and Diffraction

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Overview of Chapter 28

• Superposition and Interference

• Young’s Two-Slit Experiment

• Interference in Reflected Waves

• Diffraction

• Resolution

• Diffraction Gratings

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28-1 Superposition and Interference

If two waves occupy the same space, their amplitudes add at each point. They may interfere either constructively or destructively.

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28-1 Superposition and Interference

• Interference is only noticeable if:

- Light sources are monochromatic (so all the light has the same wavelength) and coherent (different sources maintain the same phase relationship over space and time).

• Interference will be constructive where the two waves are in phase, and destructive where they are out of phase.

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28-1 Superposition and Interference

In this illustration:

• Interference will be constructive where the path lengths differ by an integral number of wavelengths.

• Destructive where they differ by a half-odd integral number of wavelengths.

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28-1 Superposition and Interference

Two path lengths will thus interfere constructively or destructively according to the following:

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28-2 Young’s Two-Slit Experiment

In this experiment, the original light source need not be coherent; it becomes so after passing through the very narrow slits.

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28-2 Young’s Two-Slit Experiment

• If light consists of particles..

- The final screen should show two thin stripes, one corresponding to each slit.

• If light is a wave…

- Each slit serves as a new source of “wavelets”

- Final screen will show the effects of interference.

- This is called Huygens’s principle.

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28-2 Young’s Two-Slit Experiment

Pattern on the screen shows the light on the screen has alternating light and dark fringes, corresponding to constructive and destructive interference.

The path difference is given by:

Therefore, the condition for bright fringes (constructive interference) is:

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28-2 Young’s Two-Slit Experiment

The dark fringes are between the bright fringes; the condition for dark fringes is:

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28-2 Young’s Two-Slit Experiment

This diagram illustrates the numbering of the fringes.

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28-3 Interference in Reflected WavesReflected waves can interfere due to path length differences, but they can also interfere due to phase changes upon reflection.

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28-3 Interference in Reflected Waves

Constructive interference:

Destructive interference:

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28-3 Interference in Reflected WavesInterference can also occur when light refracts and reflects from both surfaces of a thin film…

•Accounts for the colors we see in oil slicks and soap bubbles.

• Now, we have not only path differences and phase changes on reflection.

• We also must account for the change in wavelength as the light travels through the film.

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28-3 Interference in Reflected WavesWavelength of light in a medium of index of refraction n:

Therefore, the condition for destructive interference, where t is the thickness of the film, is:

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28-3 Interference in Reflected WavesThe condition for constructive interference:

The rainbow of colors we see is due to the different wavelengths of light.

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28-4 Diffraction

• A wave passing through a small opening will diffract, as shown.

• Means that, after the opening, there are waves traveling in directions other than the direction of the original wave.

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28-4 Diffraction

• Diffraction is why we can hear sound even though we are not in a straight line from the source….

- Sound waves will diffract around doors, corners, and other barriers.

• Amount of diffraction depends on the wavelength, which is why we can hear around corners but not see around them.

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28-4 Diffraction

• To investigate the diffraction of light, we consider what happens when light passes through a very narrow slit.

• As the figure indicates, what we see on the screen is a single-slit diffraction pattern.

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28-4 DiffractionThis pattern is due to the difference in path length from different parts of the opening.

The first dark fringe occurs when:

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28-4 Diffraction

The second dark fringe occurs when:

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28-4 Diffraction

• In general, then, we have for the dark fringes in a single-slit interference pattern:

• The positive and negative values of m account for the symmetry of the pattern around the center.

• Diffraction fringes can be observed by holding your finger and thumb very close together (it helps not to be too farsighted!)

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28-5 Resolution

• Diffraction through a small circular aperture results in a circular pattern of fringes..

• Limits our ability to distinguish one object from another when they are very close.

• The location of the first dark fringe determines the size of the central spot Diameter of aperture

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28-5 Resolution

Rayleigh’s law (criterion) relates the size of the central spot to the limit at which two objects can be distinguished:

“If the first dark fringe of one circular diffraction pattern passes through the center of a second diffraction pattern, the two sources responsible for the patterns will appear to be a single source.”

Aperture with D…

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28-5 ResolutionOn the left, there appears to be a single source; on the right, two sources can be clearly resolved.

First dark fringes overlap First dark fringes don’t overlap

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28-6 Diffraction Gratings• A system with a large number of slits is called a diffraction grating.

• As the number of slits grows, the peaks become narrower and more intense. Here is the diffraction pattern for five slits:

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28-6 Diffraction Gratings

• The positions of the peaks are different for different wavelengths of light.

• The condition for constructive interference in a diffraction grating:

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28-6 Diffraction Gratings

X-ray diffraction is used to determine crystal structure – the spacing between crystal planes is close enough to the wavelength of the X-rays to allow diffraction patterns to be seen.

A grating spectroscope allows precise determination of wavelength:

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28-6 Diffraction Gratings• Diffraction can also be observed upon reflection from narrowly-

spaced reflective grooves; the most familiar example is the recorded side of a CD.

• Some insect wings also display reflective diffraction, especially butterfly wings.

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Answer: Destructively

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Answer : 0.18 kHz

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Answer: 623 nm

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Answer: 1.8 µm

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Answer: The resolution of an optical instrument is greater if the minimum angular separation θ is smaller.

The minimum angle θ decreases if the wavelength decreases. Therefore, greater resolution is obtained with the blue light….

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Answer: d = 0.12 nm