Single Slit (Fraunhoffer) Diffraction Question

In summary, the conversation discusses the equations for irradiance in Fraunhoffer diffraction for a rectangular aperture and a single slit. The question is whether the result for the rectangular aperture can be reduced to the result for the single slit when the height of the aperture approaches zero. The conversation also mentions a derivation for the single slit that assumes an infinitely high aperture and simplifies the integration problem. The expert concludes that when one dimension of the aperture goes to infinity, the diffraction pattern becomes a delta-function, and when it goes to zero, the pattern becomes infinitely broad.
  • #1
hbal9604@usyd
6
0
hey quick question about Fraunhoffer diffraction:

RESTULT 1

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.

QUESTION:

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)

WHEREAS

(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!

N.B.

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 across apperture of:}(e^-i(ux+vy))dxdy

where u and v are defined in the section called RESULT 1 above.
 
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  • #2
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?
 

1. What is single slit diffraction?

Single slit diffraction, also known as Fraunhoffer diffraction, is a phenomenon that occurs when light passes through a narrow slit and spreads out into a series of bright and dark fringes on a screen behind the slit.

2. What causes single slit diffraction?

Single slit diffraction is caused by the interference of light waves as they pass through the slit. The slit acts as a secondary source of light, creating a diffraction pattern on the screen.

3. How does the width of the slit affect the diffraction pattern?

The width of the slit has a direct impact on the diffraction pattern. A wider slit will produce a narrower central maximum and wider secondary maxima, while a narrower slit will produce a wider central maximum and narrower secondary maxima.

4. How is the distance between the slit and the screen related to the diffraction pattern?

The distance between the slit and the screen, also known as the screen distance, affects the spacing of the fringes in the diffraction pattern. As the screen distance increases, the fringes become wider and more spread out.

5. What is the practical application of single slit diffraction?

Single slit diffraction has many practical applications, such as in the creation of holograms, diffraction gratings, and in the study of light and wave properties. It is also used in various optical instruments, such as telescopes and microscopes, to improve resolution.

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