LECTURE 18

Instructor: Kazumi Tolich Lecture 18

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¨ Reading chapter 28.4 to 28.6 ¤ Single-slit diffraction ¤ Resolution ¤ Diffraction gratings Diffraction of waves/Demo: 1

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¨ Diffraction is bending of wave fronts due to an obstacle blocking the waves.

Plane waves Circular waves

¨ The amount of diffraction depends on the , which is why we can hear around corners but not see around them.

¨ Demo: Ripple tank Single-slit diffraction central maximum

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¨ When of wavelength � passes through a slit of width �, a diffraction pattern is formed due to the difference in path length from different parts of the opening. ¨ The wavelets going forward (� = 0) all travel the same distance to the screen and interfere constructively to produce the central bright fringe.

� Slit width � Quiz: 1

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¨ Consider the wavelets going at an such that the path length difference of the wavelet pair (1, 2) is Δr12 = �⁄2. What is the path length differences of the pairs (3, 4) and (5, 6)? A. 3�⁄2 and � B. �⁄2 and �⁄2 C. � and 3�⁄2 D. � and 2� � E. None of the above 2 Quiz: 18-1 answer

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¨ �⁄2 and �⁄2

� � � 2 2 2 Quiz: 2

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¨ Consider the wavelets going at an angle such that the path length difference of the wavelet pair (1, 2) is Δr12 = �⁄2. Do the wavelets cause a bright fringe or dark fringe? A. Bright fringe B. Dark fringe C. Neither

� 2 Quiz: 18-2 answer

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¨ Dark fringe

¨ All of the pairs have path length differences of �⁄2. So, each pair destructively interferes.

¨ Can the same technique be used to find the maxima, by choosing pairs of wavelets with path lengths that differ by λ?

¨ No. Pair-wise destructive interference works, but pair-wise constructive interference does not necessarily lead to maximum constructive interference. Single-slit diffraction/Demo: 2

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¨ The dark fringes are given by Intensity � sin � = ��, � = ±1, ±2, ±3, …

¨ Most of the light intensity is concentrated in the broad central fringe with an approximate angular width of 2 . ¨ Demo: Single slit (variable width) Diffraction of a circular

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¨ Diffraction pattern is also observed when light passes through a circular aperture.

¨ The angle � subtended by the first dark fringe is related to the wavelength � and the diameter of the opening � by

� sin � = 1.22 �

¤ the factor 1.22 arises because of the circular geometry.

Circular aperture of diameter � Various diffraction patterns/Demo: 3

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¨ Arago's Bright Spot (Poisson Spot) ¨ Point and of Needle

¨ Crossed Slits Poisson spot

¨ Knife Edge Diffraction Opaque disk ¨ Aperture Diffraction ()

“Airy disks” of a binary star viewed by a 2.56 m Rayleigh’s criterion for resolution

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¨ Rayleigh’s criterion: 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. ¨ The minimum angle subtended by the sources just resolved by Rayleigh’s criterion for resolution is � � = 1.22 � 2� � Quiz: 3

¨ The size of the smallest things that can be seen with an optical is limited by diffraction. Which of the following could help a microscopist see smaller things? Choose all that apply. A. A more powerful microscope could be used. B. The microscope could have a with a shorter . C. The microscope could have a lens with a longer focal length. D. The diameter of the lens could be smaller. E. The diameter of the lens could be larger. F. Light with a shorter wavelength could be used. G. Light with a longer wavelength could be used. Quiz: 18-3 answer

¨ The diameter of the lens could be larger. ¨ Light with a shorter wavelength could be used. ¨ � = 1.22 Example: 1

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¨ The has a mirror with a diameter of 2.4 m. For light with wavelength of 500 nm, what is the angular resolution of the Hubble? Diffraction grating/Demo: 4

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¨ 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.

¨ The condition for constructive interference in a diffraction grating:

� sin � = ��, � = 0, ±1, ±2, …

¨ Demo: multiple slit interference Quiz: 4

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¨ Monochromatic light is illuminating a diffraction grating and viewed on a far screen. Which of the following changes move/s the positions of diffraction maxima appearing on the screen? Choose all that apply. A. Change the number of rulings illuminated in the same diffraction grating. B. Change the diffraction grating with a different slit density. C. Change the frequency of the light. Quiz: 18-4 answer

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A. Change the number of rulings illuminated on the same diffraction grating.

B. Change the diffraction grating with a different slit density.

C. Change the frequency of the light.

¨ The interference maxima are given by � sin � = ��, � = 0, ±1, ±2, …

¨ Choice B changes the spacing between the slits, �.

¨ Choice C changes the wavelength of light, �.

¨ Choice A does not move the locations of maxima, but changes the intensity of maxima. ¤ The more slits a grating has, the sharper the interference maxima are. Quiz: 5

¨ White light passes through a diffraction grating and forms “rainbow” patterns on a screen behind the grating. For each rainbow: A. The side is on the left, and the side is on the right; B. The red side is on the right, and the violet side is on the left; C. The red side is closest to the center of the screen, and the violet side is farthest from the center of the screen. D. The red side is farthest from the center of the screen, and the violet side is closest to the center of the screen. Quiz: 18-5 answer/Demo: 5

¨ The red side is farthest from the center of the screen, and the violet side is closest to the center of the screen. ¨ � sin � = ��, � = 0, ±1, ±2, … ¨ The longer the wavelength �, the larger the angle � for a given diffraction grating with a fixed �.

� = −2 � = −1

Dispersion of white light