Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer

Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer Diffraction Grating with Zero Incident Angle () Is this 1 1 ? Incident plane waves (incident angle = 0) Maxima are expected when the path length difference is a multiple of a whole wavelength: a sin 0, 1, 2 … Question: What determines the direction of the m=+1 versus the m=‐1 maximum? Answer: The definition of the sign of the angle D ! We can define the sign of the angle D as follows and clarify which direction is m=+1 and m=‐1: 1 > 0 if it is counterclockwise (ccw) < 0 if it is clockwise (cw) 1 Note: We could, of course, define positive and negative exactly opposite and then m=+1 and m=‐1 would simply be exchanged. It is useful to clearly define and stick to your convention to avoid confusion. ‐ 1 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer Diffraction Grating with Non‐Zero Incident Angle () 1 0 0 The pathlength difference between adjacent slits is ∆ In our example above: a sin 0 a sin 0 Maxima will occur when a sin sin 0, 1, 2 … For 0 sin sin " "! 0 0 0 ‐ 2 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer Reflection Gratings with Incident Angle Is this 1 1 ? Δsin Maxima are expected when the path length difference is a multiple of a whole wavelength: a sin 0, 1, 2 … Definitions: ccw = positive cw = negative In the example above: m = positive because D = positive ‐ 3 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer Reflection Gratings with Incident Angle dsin 0 0 isin In this situation, both i and d are positive because both angles and are positive. ∆ sin sin sin sin Maxima occur when sin sin 0,1,2… ‐ 4 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer isin dsin 0 0 In this situation: sin 0 and sin 0 Maxima occur when sin sin 0,1,2… 0 maximum occurs when 0 maxima occur under the following conditions: 0 |||| OR 0 |||| 1 0 1 || || 0 ‐ 5 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer What is the Benefit of a Blaze Angle? Grating with a blaze angle of 0 (no blazing): Specular reflection: || || The direction of specular reflection is also the direction for the m=0 diffraction maximum. If and then sin sin sin sin 0 0 Specular reflection is in the direction of the m=0 diffraction maximum. Most of the power of the light gets diffracted into the m=0 maximum, which is usually not of interest because all the different wavelengths have their m=0 maximum in that same direction. Relatively little power goes into the spectroscopically interesting higher order diffraction maxima which split the maxima according to wavelegth. ‐ 6 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer Grating with a blaze angle : Incident beam Surface Normal Θ Θ Θ 2Θ Θ Θ Direction of specular reflection Optical Axis 0 diffraction maximum Surface normal does not coincide with the optical axis. Most of the optical power gets diffracted in the direction of specular reflection, which is NOT the direction of the m=0 diffraction maximum. The blaze angle can be designed such that for a particular wavelength the direction of specular reflection is also the direction of a diffraction maximum with m0. In the example above we can design the angle such that for wavelength the m=+1 diffraction maximum goes in the direction of specular reflection: Θ Θ 2Θ For the 1 maximum: sin sin sin 2γ D sin 2Θ 22 2 Θ Solution (see Appendix): sin γ sin ; 2 ‐ 7 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer Diffraction gratings are typically blazed with a blaze angle for a particular wavelength and line spacing a to fulfill the condition: ; 2 The best angle of incidence is the blaze angle and the m=+1 diffraction goes right back to where it came from. Appendix Suppose you have a blazed grating with parameters a (line spacing) and (blaze angle) and you shine on it light with wavelength : What is the ideal angle of incidence to get the m=+1 diffraction maximum as close as possible to the direction of specular reflection? Incident beam Surface Normal Direction of specular reflection Direction of m=+1 maximum Optical Axis 2 0 diffraction maximum Direction of specular reflection: 2 differ by and angle Direction of m=+1 diffraction: ‐ 8 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer m = 1: sin sin sin 2 2 sin Ideally 0 , but that is not necessarily possible. (Here is an example of a condition where 0: 2 sin 0 sin sin sin ) Optimizing for given , , : cos 1 2 sin 1 Note: √1 1 sin cos 1 0 1 sin cos 1 sin cos I 1 sin cos I 1 2 sin 0 2 sin 2sin sin 2 2 Since sin sin sin sin sin 2 ‐ 9 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer What is under these conditions? 2 2 2 sin 2 22 2 2 2 2 2 Can be small for 0 By the way: 0 for What is the ideal angle of incidence to get the m=‐1 diffraction maximum as close as possible to the direction of specular reflection? m = ‐1: sin sin sin 2 2 sin Optimizing for given , , : cos 2 sin 1 1 sin cos 1 0 1 sin cos 1 sin cos I 1 sin ‐ 10 ‐ Physics 6775 – G. Laicher Spring 2012 Lab 7 Part 1: Background Lecture ‐ Spectrometer cos I 1 2 sin 0 2 sin sin 2 2 Since sin sin sin sin sin 2 2 What is under these conditions? 2 2 2 positive positive 2 22 2 2 2 2 2 Is large !! You would have to flip the grating so that γ ‐γ in order to be able to make ε small. ‐ 11 ‐ .

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