Chapter 36

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# Chapter 36 - PowerPoint PPT Presentation

Chapter 36. Diffraction. Goals for Chapter 36. To see how a sharp edge or an aperture affect light To analyze single-slit diffraction and calculate the intensity of the light To investigate the effect on light of many closely spaced slits To learn how scientists use diffraction gratings

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## Chapter 36

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Presentation Transcript
1. Chapter 36 Diffraction

2. Goals for Chapter 36 • To see how a sharp edge or an aperture affect light • To analyze single-slit diffraction and calculate the intensity of the light • To investigate the effect on light of many closely spaced slits • To learn how scientists use diffraction gratings • To see what x-ray diffraction tells us about crystals • To learn how diffraction places limits on the resolution of a telescope

3. Introduction • How can we use coherent light to visually see the difference in pit density on CDs, DVDs and Blu-Ray disks? • Why does light from a point source form light and dark fringes when it shines on a razor blade? • We will continue our exploration of the wave nature of light with diffraction. • And we will see how to form three-dimensional images using a hologram.

4. 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.

5. Diffraction and Huygen’s Principle • Huygens’s principle can be used to analyze diffraction. • 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.

6. 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). • The narrower the slit, the broader the diffraction pattern.

7. 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.

8. Locating the dark fringes • Follow the single-slit diffraction discussion in the text. • Figure 36.5 below shows the geometry for Fraunhofer diffraction.

9. An example of single-slit diffraction • Figure 36.6 (bottom left) is a photograph of a Fraunhofer pattern of a single horizontal slit. • Example 36.1: You pass 633-nm light through a narrow slit and observe the diffraction pattern on a screen 6.0 m away. The distance at the screen between the center and the first minima on either side is 32 mm long. How wide is the slit?

10. Intensity in the single-slit pattern • Follow the text discussion of the intensity in the single-slit pattern using the phasor diagrams in Figure 36.8 below.

11. Movie Showing Development of the Pattern • In the movie, the blue “circle” represents the E vectors from each section of the slit, as we move along different positions on the screen. • The green line represents the resultant vector Ep from adding all of the individual vectors. • The red line is the x-component of Ep, or Epcosf. • The cyan line merely traces out Epcosf along the screen.

12. Quantitative Intensity in the single-slit pattern • Follow the text discussion of the intensity in the single-slit pattern using the phasor diagrams in Figure 36.8 below. • The angle b is the phase angle of the ray from the top of the slit, while the phase angle from the bottom of the slit is 0. The vectors lie along a circle whose center is at C, so Ep is a chord of the circle. The arc length E0 is subtended by this same angle b, so the radius of the circle is E0/b. • From the diagram, • Since • We have (sinc function)

13. Intensity maxima in a single-slit pattern • Figure 36.9 at the right shows the intensity versus angle in a single-slit diffraction pattern. • The minima occur when b is a multiple of 2p, i.e. at • The location of the maxima are found by taking the derivative of • and setting it to zero. Surprisingly, these are not precisely where • In fact, there are no maximum for m = 0 in this expression. The central maximum is wider than the others, and occurs at q = 0. • Using these approximate values of b in the intensity, we find

14. Width of the single-slit pattern • The single-slit diffraction pattern depends on the ratio of the slit width a to the wavelength . • Example 36.2: (a) The intensity at the center of a single-slit diffraction pattern is I0. What is the intensity at a point in the pattern where there is a 66-radian phase difference between wavelets from the two edges of the slit? (b) If this point is 7 degrees from the central maximum, how many wavelengths across is the slit? • (a) • (b)

15. Two slits of finite width • When we discussed two-slit interference in Chapter 35, we ignored the width of each slit. When we demonstrated it, however, we saw clearly the effect of the slit widths. • The overall pattern of two finite-width slits is the product of the two patterns, i.e. • where

16. Several slits • In Figure 36.13 below, a lens is used to give a Fraunhofer pattern on a nearby screen. It’s function is to allow the pattern to be seen nearby, without having the screen really distant. • The phasor diagrams show the electric vectors from each slit at different screen locations.

17. Interference pattern of several slits • The figure below shows the interference pattern for 2, 8, and 16 equally spaced narrow slits. • By making the slits really close together, the maxima become more separated. If the light falling on the slits contains more than one wavelength (color), there will be more than one pattern, separated more or less according to wavelength, although all colors have a maximum at m = 0. • This means that the different orders make rainbows—separating wavelengths into a spectrum, with the separation being greater for greater order m.

18. The diffraction grating • A diffraction grating is an array of a large number of slits having the same width and equal spacing. The intensity maxima occur at • Example 36.4: The wavelengths of the visible spectrum are approximately 380 nm (violet) to 750 nm (red). (a) Find the angular limits of the first-order visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating. (b) Do the first order and second order spectra overlap? What about the 2nd and 3rd orders? • (a) distance between slits is • Violet light for 1st order occurs at • Red light for 1st order occurs at • (b) recalculate for m = 2 and m = 3. The 2nd-order spectrum extends from 27.1-63.9° while the 3rd order is from 43-90.

19. Grating spectrographs • A diffraction grating can be used to disperse light into a spectrum. • The greater the number of slits, the better the resolution. • Figure 36.18(a) below shows our sun in visible light, and in (b) dispersed into a spectrum by a diffraction grating. See description of Eschelle spectrograph: http://www.vikdhillon.staff.shef.ac.uk/teaching/phy217/instruments/phy217_inst_echelle.html

20. Diagram of a grating spectrograph • Figure 36.19 below shows a diagram of a diffraction-grating spectrograph for use in astronomy.

21. X-ray diffraction • When x rays pass through a crystal, the crystal behaves like a diffraction grating, causing x-ray diffraction. Figure 36.20 below illustrates this phenomenon.

22. A simple model of x-ray diffraction • Follow the text analysis using Figure 36.22 below. • The Bragg condition for constructive interference is 2d sin = m. • Follow Example 36.5.

23. Circular apertures • An aperture of any shape forms a diffraction pattern. • Figures 36.25 and 36.26 below illustrate diffraction by a circular aperture. The airy disk is the central bright spot. • The first dark ring occurs at an angle given by sin1 = 1.22 /D.

24. Diffraction and image formation • Diffraction limits the resolution of optical equipment, such as telescopes. • The larger the aperture, the better the resolution. Figure 36.27 (right) illustrates this effect.

25. Bigger telescope, better resolution • Because of diffraction, large-diameter telescopes, such as the VLA radiotelescope below, give sharper images than small ones. • Follow Example 36.6.

26. What is holography? • By using a beam splitter and mirrors, coherent laser light illuminates an object from different perspectives. Interference effects provide the depth that makes a three-dimensional image from two-dimensional views. Figure 36.28 below illustrates this process.

27. How does holography work? • Follow the text analysis using Figure 36.29 below.

28. An example of holography • Figure 36.32 below shows photographs of a holographic image from two different angles, showing the changing perspective.