LECTURE 17 DIFFRACTION OFF CIRCULAR APERTURE & PHASORS

Instructor: Kazumi Tolich Lecture 17

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¨ Reading chapter 33-5 to 33-7 ¤ Fraunhofer and Fresnel diffraction ¤ Diffraction and resolution ¤ Phasors n Addition of two harmonic waves n Interference pattern from multiple sources n Single slit diffraction pattern n Interference-diffraction pattern Fraunhofer diffraction of a circular aperture

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¨ Diffraction pattern is also observed when light passes through a circular aperture instead of a vertical slit.

¨ The angle θ subtended by the first diffraction minimum is related to the wavelength and the diameter of the opening D by

¤ the factor 1.22 arises because of the circular geometry.

Circular aperture of diameter D Fraunhofer and Fresnel diffraction patterns

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¨ Diffraction patterns formed when the viewing screen is far from the apertures so that the rays from the apertures are parallel are called Fraunhofer diffraction patterns. ¤ They can be observed using a lens to focus parallel rays on a viewing screen placed in the focal plane of the lens.

¨ The diffraction pattern observed at any distance from an aperture or an obstacle is called a Fresnel diffraction pattern. Fresnel diffraction patterns

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¨ The patterns of a circular aperture and a opaque disk are complements of each other.

¨ Poisson spot at the center of the pattern caused by the constructive interference of the light waves diffracted from the edge of the disk.

Poisson spot

Opaque disk Circular aperture Fresnel diffraction patterns 2

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¨ Fresnel Diffraction Patters of a rectangular aperture (left) and a straightedge (right).

Straightedge

Rectangular aperture Demo: 1

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¨ Arago's Bright Spot (Poisson Spot) ¤ Demonstration of Poisson spot

¨ Point and Eye of Needle ¤ Demonstration of diffraction due to a needle

¨ Crossed Slits ¤ Two dimensional diffraction pattern

¨ Knife Edge Diffraction “Air y disks” of a ¨ Aperture Diffraction (Airy Disk) binary star viewed by a 2.56 m telescope ¤ Demonstration of diffraction due to circular aperture Rayleigh’s criterion for resolution

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¨ When the first minimum of the diffraction pattern of one source falls on the central maximum of the other source, these objects are just resolved by Rayleigh’s criterion for resolution. ¨ The critical angle subtended by the sources just resolved by Rayleigh’s criterion for resolution is

2αc αc Quiz: 1 Example 1

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¨ The Hubble Space Telescope 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? Example 2

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¨ How far from the grains of red sand must you be to position yourself just at the limit of resolving the grains if your pupil diameter is 1.5 mm, the grains are spherical with radius 50.0 µm, and the light from the grains has wavelength 650 nm? If the grains were blue, and the light from them had wavelength 400 nm, would the answer be larger or smaller? Addition of two harmonic waves and phosors

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¨ Addition of two waves with different phases can be graphically expressed with phasors.

¨ Each wave function is represented by the y component of a phasor.

¨ � = � sin �, where � = ��. ¨ � = � sin � + � ¨ � + � = � sin � + � Three slit interference

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¨ Assume the rays hitting the screen are parallel. ¨ Interference patters appear due to the path length differences of the rays.

¨ The wave at a point on the screen, P, is the sum of the three waves.

¨ � = � sin � 1 ¨ � = � sin � + �

¨ � = � sin � + 2� 2 ¨ � = 2� 3 Quiz: 2

14 Multi-slit interference patterns

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¨ The principal maxima occur at the same θ as for two slit interference.

¨ The more slits there are, the brighter and narrower the principal maxima.

¨ The first minimum occurs at � = 360°⁄� (N phasors form a closed polygon of N sides).

¨ There are N - 2 secondary maxima between each pair of principal maxima.

¨ There are N - 1 zeros between each pair of principal maxima.

5 1 I / 4 N I

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θ (radians) Diffraction patterns

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¨ The slit of width a is divided into N (N >> 1) equal intervals.

¨ The rays from the sources to a point P on the far screen d are parallel. To screen at point P ¨ The diffraction pattern arises from the path length difference between any two adjacent sources resulting in the phase difference given by � sin � � = 2� � ¨ Let A0 denote the amplitude due to a single source, then � = � sin � + �� Diffraction pattern and phasors

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Intensity I0 Interference-diffraction pattern

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¨ However, in a two-slit system interference and d diffraction should be present together. a a ¨ Interference only: Interference only � = 4� cos , where � = ¤ � is the intensity from one slit. ¨ Diffraction only: q (degrees) ⁄ � = � , where � = Diffraction only ⁄ ¤ � is the intensity at the center. q (degrees) ¨ Combined: Both sin �⁄2 � � = 4� cos �⁄2 2 q (degrees) ¤ � is the intensity at the center from one slit. Demo: 2

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¨ Multiple Slit Interference/Diffraction Example: 3

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¨ Light that has a wavelength equal to 550 nm illuminates two slits that both have widths equal to 0.030 mm and separations equal to 0.18 mm. This creates 11 fringes in the central diffraction maximum. What is the ratio of the intensity of the third interference maximum to the intensity of the center interference maximum? Imax I3