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Lecture 40: MON 27 APR

Physics 2102 Jonathan Dowling. Lecture 40: MON 27 APR. Ch. 36: Diffraction. Things You Should Learn from This Lecture.

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Lecture 40: MON 27 APR

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  1. Physics 2102 Jonathan Dowling Lecture 40: MON 27 APR Ch. 36: Diffraction

  2. Things You Should Learn from This Lecture • When light passes through a small slit, is spreads out and produces a diffraction pattern, showing a principal peak with subsidiary maxima and minima of decreasing intensity. The primary diffraction maximum is twice as wide as the secondary maxima. • We can use Huygens’ Principle to find the positions of the diffraction minima by subdividing the aperture, giving qmin= ±p l/a, p = 1, 2, 3, ... . • Calculating the complete diffraction pattern takes more algebra, and gives Iq=I0[sin(a)/a]2, where a = p a sin(q)/l. • To predict the interference pattern of a multi-slit system, we must combine interference and diffraction effects.

  3. Single Slit Diffraction When light goes through a narrow slit, it spreads out to form a diffraction pattern.

  4. Analyzing Single Slit Diffraction For an open slit of width a, subdivide the opening into segments and imagine a Hyugen wavelet originating from the center of each segment. The wavelets going forward (q=0) all travel the same distance to the screen and interfere constructively to produce the central maximum. Now consider the wavelets going at an angle such that l = a sin q @a q. The wavelet pair (1, 2) has a path length difference Dr12 = l/2, and therefore will cancel. The same is true of wavelet pairs (3,4), (5,6), etc. Moreover, if the aperture is divided into p sub-parts, this procedure can be applied to each sub-part. This procedure locates all of the dark fringes.

  5. Conditions for Diffraction Minima

  6. Pairing and Interference Can the same technique be used to find the maxima, by choosing pairs of wavelets with path lengths that differ by l? No. Pair-wise destructive interference works, but pair-wise constructive interference does not necessarily lead to maximum constructive interference. Below is an example demonstrating this.

  7. Calculating theDiffraction Pattern We can represent the light through the aperture as a chain of phasors that “bends” and “curls” as the phase Db between adjacent phasors increases. b is the angle between the first and the last phasor.

  8. Calculating theDiffraction Pattern (2)

  9. l = 633 nm a = 0.25 mm 0.5 mm 1 mm 2 mm Blowup q (radians) Diffraction Patterns The wider the slit opening a, or the smaller the wavelength , the narrower the diffraction pattern.

  10. Radar: The Smaller The Wavelength the Better The Targeting Resolution Laser: =1 m  Ka-band: =1cm X-band: =10cm K-band: =2cm

  11. Angles of the Secondary Maxima The diffraction minima are precisely at the angles wheresin q = p l/a and a = pp(so that sin a=0). However, the diffraction maxima are not quite at the angles where sin q= (p+½) l/aand a= (p+½)p (so that |sin a|=1). l = 633 nm a = 0.2 mm 1 2 3 4 5 q(radians) To find the maxima, one must look near sin q= (p+½) l/a, for places where the slope of the diffraction pattern goes to zero, i.e., whered[(sin a/a)2]/dq = 0. This is a transcendental equation that must be solved numerically. The table gives the qMax solutions. Note that qMax < (p+½)l/a.

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