Chapter 36 Diffraction Power Point Lectures for University

Chapter 36 Diffraction Power. Point® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Copyright © 2012 Pearson Education Inc.

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 Copyright © 2012 Pearson Education Inc.

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 threedimensional images using a hologram. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

Locating the dark fringes • Follow the single-slit diffraction discussion in the text. • Figure 36. 5 below shows the geometry for Fraunhofer diffraction. Copyright © 2012 Pearson Education Inc.

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? Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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 xcomponent of Ep, or Epcosf. • The cyan line merely traces out Epcosf along the screen. Copyright © 2012 Pearson Education Inc.

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 E 0 is subtended by this same angle b, so the radius of the circle is E 0/b. • From the diagram, • Since • We have (sinc function) Copyright © 2012 Pearson Education Inc.

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 2 p, 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 = 0. • Using these approximate values of b in the intensity, we find Copyright © 2012 Pearson Education Inc.

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 I 0. 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) Copyright © 2012 Pearson Education Inc.

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 Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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 firstorder 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 2 nd and 3 rd orders? • (a) distance between slits is Violet light for 1 st order occurs at Red light for 1 st order occurs at • (b) recalculate for m = 2 and m = 3. The 2 nd-order spectrum extends from 27. 1 -63. 9° while the 3 rd order is from 43 -90. Copyright © 2012 Pearson Education Inc.

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/phy 217/instruments/phy 217_inst_echelle. html Copyright © 2012 Pearson Education Inc.

Diagram of a grating spectrograph • Figure 36. 19 below shows a diagram of a diffractiongrating spectrograph for use in astronomy. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

A simple model of x-ray diffraction • Follow the text analysis using Figure 36. 22 below. • The Bragg condition for constructive interference is 2 d sin = m. • Follow Example 36. 5. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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. Copyright © 2012 Pearson Education Inc.

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 threedimensional image from two-dimensional views. Figure 36. 28 below illustrates this process. Copyright © 2012 Pearson Education Inc.

How does holography work? • Follow the text analysis using Figure 36. 29 below. Copyright © 2012 Pearson Education Inc.

An example of holography • Figure 36. 32 below shows photographs of a holographic image from two different angles, showing the changing perspective. Copyright © 2012 Pearson Education Inc.