 # Chapter 35 Interference (cont.).

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Chapter 35 Interference (cont.)

Interference in thin films
Figure (right) shows why thin-film interference occurs, with an illustration. Figure (below) shows interference of an air wedge.

Phase shifts during reflection
Follow the text analysis of thin-film interference and phase shifts during reflection. Use Figure below.

Q35.6 An air wedge separates two glass plates as shown. Light of wavelength l strikes the upper plate at normal incidence. At a point where the air wedge has thickness t, you will see a bright fringe if t equals l/2. 3l/4. C. . D. either A. or C. E. any of A., B., or C. Answer: B

A35.6 An air wedge separates two glass plates as shown. Light of wavelength l strikes the upper plate at normal incidence. At a point where the air wedge has thickness t, you will see a bright fringe if t equals l/2. 3l/4. C. . D. either A. or C. E. any of A., B., or C.

Newton’s rings Figure below illustrates the interference rings (called Newton’s rings) resulting from an air film under a lens.

Using interference fringes to test a lens
The lens to be tested is placed on top of the master lens. If the two surfaces do not match, Newton’s rings will appear, as in Figure at the right.

Chapter 36 Diffraction

Diffraction According to geometric optics, a light source shining on an object in front of a screen should cast a sharp shadow. Surprisingly, this does not occur because of diffraction.

Diffraction and Huygen’s Principle
Fresnel diffraction: Source, screen, and obstacle are close together. Fraunhofer diffraction: Source, screen, and obstacle are far apart. Figure 36.2 below shows the diffraction pattern of a razor blade.

Diffraction from a single slit
In Figure 36.3 below, the prediction of geometric optics in (a) does not occur. Instead, a diffraction pattern is produced, as in (b).

Fresnel and Fraunhofer diffraction by a single slit
Figure 36.4 below shows Fresnel (near-field) and Frauenhofer (far-field) diffraction for a single slit.

Locating the dark fringes
Figure 36.5 below shows the geometry for Fraunhofer diffraction.

An example of single-slit diffraction
Figure 36.6 (bottom left) is a photograph of a Fraunhofer pattern of a single horizontal slit.

Intensity maxima in a single-slit pattern
Figure 36.9 at the right shows the intensity versus angle in a single-slit diffraction pattern. Part (b) is photograph of the diffraction of water waves.

Width of the single-slit pattern
The single-slit diffraction pattern depends on the ratio of the slit width a to the wavelength . (Figure below.)

A. Double the slit width a and double the wavelength l.
Q36.1 Light of wavelength l passes through a single slit of width a. The diffraction pattern is observed on a screen that is very far from from the slit. Which of the following will give the greatest increase in the angular width of the central diffraction maximum? A. Double the slit width a and double the wavelength l. B. Double the slit width a and halve the wavelength l. C. Halve the slit width a and double the wavelength l. D. Halve the slit width a and halve the wavelength l. Answer: C

A36.1 Light of wavelength l passes through a single slit of width a. The diffraction pattern is observed on a screen that is very far from from the slit. Which of the following will give the greatest increase in the angular width of the central diffraction maximum? A. Double the slit width a and double the wavelength l. B. Double the slit width a and halve the wavelength l. C. Halve the slit width a and double the wavelength l. D. Halve the slit width a and halve the wavelength l.

What is the maximum slit width a for which this occurs?
Q36.2 In a single-slit diffraction experiment with waves of wavelength l, there will be no intensity minima (that is, no dark fringes) if the slit width is small enough. What is the maximum slit width a for which this occurs? A. a = l/2 B. a = l C. a = 2l D. The answer depends on the distance from the slit to the screen on which the diffraction pattern is viewed. Answer: B

A36.2 In a single-slit diffraction experiment with waves of wavelength l, there will be no intensity minima (that is, no dark fringes) if the slit width is small enough. What is the maximum slit width a for which this occurs? A. a = l/2 B. a = l C. a = 2l D. The answer depends on the distance from the slit to the screen on which the diffraction pattern is viewed.

Two slits of finite width
For slits extremely narrow, behaves very close to ideal case from previous chapter For wider slits, behaves like a combination of single-slit diffraction and double-slit interference.

Interference pattern of several slits
Figure below shows the interference pattern for 2, 8, and 16 equally spaced narrow slits.

A. The bright areas move farther apart.
Q36.3 In Young’s experiment, coherent light passing through two slits separated by a distance d produces a pattern of dark and bright areas on a distant screen. If instead you use 10 slits, each the same distance d from its neighbor, how does the pattern change? A. The bright areas move farther apart. B. The bright areas move closer together. C. The spacing between bright areas remains the same, but the bright areas become narrower. D. The spacing between bright areas remains the same, but the bright areas become broader. Answer: C

A36.3 In Young’s experiment, coherent light passing through two slits separated by a distance d produces a pattern of dark and bright areas on a distant screen. If instead you use 10 slits, each the same distance d from its neighbor, how does the pattern change? A. The bright areas move farther apart. B. The bright areas move closer together. C. The spacing between bright areas remains the same, but the bright areas become narrower. D. The spacing between bright areas remains the same, but the bright areas become broader.