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Single Slit (Fraunhoffer) Diffraction Question

  1. Apr 6, 2010 #1
    hey quick question about Fraunhoffer diffraction:


    for a rectanguar apperture, we have the following equation for irradiance I:

    I = I(0)(sin(B)/B)^2(sin(A)/A)^2 = I(0)[sinc(B)]^2[sinc(A)]^2

    where, B = bu/2; A = av/2
    where b = apperture breadth (x-axis); a = apperture height (y-axis)
    and u = ksin(theta_x) v = ksin(theta_y)

    and where I(0) is the irradiance evaluated along the central axis between the apperture and the screen, ie at theta_x = 0, and theta_y = 0.

    RESULT 2

    Now, I also have the equation for single slit diffraction:

    I = I(0)(sin(B)/B)^2 = I(0)[sinc(B)]^2

    also here, B = bu/2;
    where b = apperture breadth (x-axis);
    and u = ksin(theta_x)

    where I(0) is the irradiance evaluated along the central axis between the apperture and the screen, ie where theta_x = 0.


    RESULT 1 and 2 can be derived independently of each other using double intregral formulas. Now my question is as follows:

    is it satisfactory to think of RESULT 1 reducing to RESULT 2 in the limit that (say) the height (y-axis dimension) of the rectuangular apperture approaches zero?

    In this case, the apperture height, a ----> 0
    thus A-----> 0
    thus sinc(A)----> 1
    thus (from RESULT 1) I----> I(0)(sin(B)/B)^2 x 1 = I(0)sinc(B)^2 as required.

    This seems to make sense to me. and what I get is a result where by a single (infinitely thin) slit of length b in the x-dimenstion, produces a diffraction pattern on the screen in teh SAME dimension (ie the x-dimension).

    WHY I ASK.........is that in one form of the derivation of RESULT 2 for the single slit I have come across, the following is stated:

    'consider a slit with width b and assume the hight is infinite'.

    FYI: In this derivation, the width is still along the x-axis, and the height is along the y-axis.

    the working then proceeds to claim that SINCE we have assumed the slit is infinitely high, we need not consider integrating across the y-variable, and so the integration problem is simplifed to a single integration from -b/2 to +b/2 in the x-variable. RESULT 2 is then achieved.

    YOU MAY HAVE SEEN MY PROBLEM AT THIS POINT (or maybe I have missed the point):

    (i) in the derivation I just related to you, we end up with RESULT 2, but we do so by applying the fraunhoffer double integral to a slit of finite width b (along the x-axis) and infinite height a (along the y-axis)


    (ii) when I make RESULT 1 reduce to RESULT 2, I do so by assuming that our slit has finite width b (along the x-axis) but ZERO hieght (a = 0) along the y-axis.

    Could you please comment on my thoughts, and point out my error if one exists. THANK YOU!


    when I refer to the fraunhoffer double integral, I refer to the following:

    Electric Field At point P on the Screen =

    constant x {double integral accross apperture of:}(e^-i(ux+vy))dxdy

    where u and v are defined in the section called RESULT 1 above.
  2. jcsd
  3. Apr 6, 2010 #2

    Andy Resnick

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    Education Advisor

    If I understand your post, going from a rectangular aperture to a slit means the dimension in one axis goes to *infinity*, and so the diffraction along that axis goes to a delta-function. If one dimension of the aperture goes to zero (a delta function), then the diffraction pattern becomes infinitely broad.

    Or did I not read something correctly?
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