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# Lect04 - Applications of Interference and Diffraction I0...

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α=2α c 0 I I 0 y Sum 2I 0 0 0 I I 0 Sum 2I 0 0 α=α c 0 I I 0 Sum 2I 0 0 α=α c /3 y y Applications of Interference and Diffraction:

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Overview Overview c Circular Diffraction ( foreshadowing of quantum uncertainty) c Angular resolution (Rayleigh’s criterion) c Minimum spot size c Interferometers c Michelson c Applications
Diffraction Diffraction - - limited Optics limited Optics c Diffraction has important implications for optical instruments Lens-making is a craft. Even for a perfectly designed lens, however, the image of a point source will be a little “blurry” due to diffraction in passing through the circular aperture of the lens. The image of a point source through a circular aperture is like a single-slit diffraction pattern. But note the difference: The amount of ‘smearing’ of the image is determined by size of the aperture D, and wavelength of incident light, λ. λ. a λ θ 0 Slit 0 1.22 D λ θ ≈ Circular aperture Image plane 0 0 I I 0 θ o θ D Image plane Point object

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Transmission of light through slits and circular apertures Transmission of light through slits and circular apertures Monochromatic light source at a great distance, or a laser. Object at any distance: Slit, width a Lens, diameter D Pinhole, diameter D θ -λ/ a 0λ/ a 0 I I 0 θ -1.22λ/ D 01.22λ/ D 0 I I 0 θ -1.22λ/ D 01.22λ/ D 0 I I 0 Observation screen: Image Plane: Observation screen: Laser with pinholes Circular-aperture diffraction pattern =“the Airy disk”. Central lobe contains 84% of power.
In 1985, a laser beam with a wavelength of λ = 500 nm was fired from the earth and reflected off the space shuttle Discovery, in orbit at a distance of L = 350 km away from the laser. If the (circular) aperture of the laser was D = 4.7 cm , what was the beam diameter d at the space shuttle? Exercise 1: Exercise 1: Expansion of a Laser beam D d

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In 1985, a laser beam with a wavelength of λ = 500 nm was fired from the earth and reflected off the space shuttle Discovery, in orbit at a distance of L = 350 km away from the laser. If the (circular) aperture of the laser was D = 4.7 cm , what was the beam diameter d at the space shuttle? Exercise 1: Exercise 1: Expansion of a Laser beam - Solution Solution D d Half-angle-width of diffraction maximum: 9 5 2 500 10 1.22 1.22 1.3 10 radians 4.7 10 o D λ θ - - - × = = = × × 5 3 o d 2 L 2(1.3 10 )(350 10 m) 9.1 m - ≈ θ × = × × = 84% of power is in central lobe.
λ = 500 nm was fired from the earth and reflected off the space shuttle Discovery, in orbit at a distance of L = 350 km away from the laser. To make a smaller spot on the shuttle, what should we do to the beam diameter at the source? a. reduce it

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Lect04 - Applications of Interference and Diffraction I0...

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