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Fraunhofer Diffraction Pattern Ratio of Power Densities

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Homework Statement


Find an approximate expression for the ratio of the power densities at the principal maximum to that at the first secondary maximum on either side, in the Fraunhofer diffraction pattern of an N-slit multiple aperture. Assume that the slits are much narrower than their separation, and that the peak of the secondary maximum occurs halfway between the first and second zeros of the pattern (not exactly true).

Homework Equations


The width of the central diffraction peak

[1] Diffraction minima: nλ=b sinθ

[2] Central interference fringe: pλ=a sinθ

where
a: width of slit
b: width of portion between slits
θ: angle of diffraction
λ: wavelength of light
p & n: both integers

The Attempt at a Solution



Dividing [2] by [1]

pλ / mλ = p/n

a sinθ / b sinθ = a/b

so a/b = p/n

∴ 2(a/b) = ratio of widths of the central diffraction peak to the central interference fringe

Can this also be extended to the ratio of power densities of the Fraunhofer diffraction pattern? I can't really think of any other way to answer the question. Any help would be much appreciated.



 

Answers and Replies

  • #2
Charles Link
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You need the formula ## I=I_o \frac{\sin^2(\frac{N \phi}{2})}{\sin^2(\frac{\phi}{2})} ## where ## \phi=\frac{2 \pi \, d \, \sin{\theta}}{\lambda} ##. ## \\ ## And a hint here: the intensity of the primary maxima including the central maximum is found as the limit when the denominator of the formula goes to zero.
 
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  • #3
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I'm assuming the denominator goes to zero because the angle, sinθ, equals 0, thus reducing Φ and I to zero?
 
  • #4
Charles Link
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I'm assuming the denominator goes to zero because the angle, sinθ, equals 0, thus reducing Φ and I to zero?
See the additions/hint of post 2. ## \\ ## ## I=N^2 I_o ## at the primary maxima, where ## m \lambda =d \sin{\theta} ##. (Note: ## \frac{\phi}{2}=m \pi ## at the primary maxima, and that's what makes the denominator zero). ## \\ ## And yes, the denominator is zero at the primary maxima. It is extremely useful, but it takes a little practice to learn how to use the formula of post 1. You need to take the limit as ## \frac{\phi}{2} \rightarrow m \pi ##. ## \\ ## e.g. Let ## \frac{\phi}{2}=m \pi+\delta ##.
 
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  • #5
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Where have you got the original equation from? My notes don't have it
 
  • #6
Charles Link
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Where have you got the original equation from? My notes don't have it
See Hecht and Zajac Optics. I think it is also in the second book of Halliday and Resnick that covers E&M.## \\ ## I specialized somewhat in diffraction grating type spectroscopy, ( =with spectrometers that use a diffraction grating, (as opposed to a prism), to get the dispersion), so that is one of a number of formulas that I have at my fingertips=It's used quite frequently in diffraction grating spectroscopy.
 
  • #7
Charles Link
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See also http://www.physics.louisville.edu/cldavis/phys299/notes/lo_msgratings.html These notes do not derive the formula for N-slit interference, (they just state the result), and included in this formula here is the diffraction factor that comes from the diffraction of a single slit. For narrow slits, this diffraction factor is 1, and you just get the interference part. His ## \gamma=\frac{\phi}{2} ##. ## \\ ## And note: To work the problem you have, the secondary maximum next to the ## m ##th primary maximum will occur when ## \frac{N \phi}{2}=Nm \pi \pm \frac{\pi}{2} ##. (Edit: Mistake here: See post 29). This will make the numerator equal to 1. You just need to solve for ## \frac{\phi}{2} ##, and plug it in, and use the trig formula for ## \sin(\theta+\phi) ##.
 
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  • #8
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I'm a little confused with the terminology between primary maxima and central maxima
 
  • #9
Charles Link
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I'm a little confused with the terminology between primary maxima and central maxima
The central maximum has ## m=0 ##. My mention of it was perhaps unnecessary in post 2. ## \\ ## Additional item: Primary maxima occur when ## m \lambda=d \sin{\theta} ##.(The path difference ## \Delta =d \sin{\theta} ## (to a point at location ## \theta ## in the far-field) between adjacent slits is an integer number of wavelengths). ## \\ ## (Note: This is the same thing as ## \phi=(2 \pi)(m) ##). ## \\ ## When this condition is met, adjacent slits constructively interfere. If adjacent slits are ## (2 \pi) ( m ) ## out of phase with each other, (as observed at a location ## \theta ## in the far-field), all of the ## N ## the slits in the grating will constructively interfere.
 
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  • #10
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I'm really confused with this question

## I=I_o \frac{\sin^2(\frac{N \phi}{2})}{\sin^2(\frac{\phi}{2})} ## where ## \phi=\frac{2 \pi \, d \, \sin{\theta}}{\lambda} ##. ## \\ ##

This is the power density due to N slits. We want a ratio of the power density between the central maximum to the secondary maximum. I'm trying to do this by first calculating the power density at the central maximum, and then the power density at the secondary maximum.

## \phi=\frac{2 \pi \, d \, \sin{\theta}}{\lambda} ##

dsinθ=mλ

If the central maximum has m=0, then sinθ=0

But if m=0, then Φ=0, and thus I=0 ...

Power density at the central maximum = 0

Secondary maximum m=1, Φ=2π

Intensity for secondary maximum is then:

I = I0 [ sin2(Nπ) / sin2(π) ]

I'm not sure what I'm missing here...
 
  • #11
Charles Link
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I'm really confused with this question

## I=I_o \frac{\sin^2(\frac{N \phi}{2})}{\sin^2(\frac{\phi}{2})} ## where ## \phi=\frac{2 \pi \, d \, \sin{\theta}}{\lambda} ##. ## \\ ##

This is the power density due to N slits. We want a ratio of the power density between the central maximum to the secondary maximum. I'm trying to do this by first calculating the power density at the central maximum, and then the power density at the secondary maximum.

## \phi=\frac{2 \pi \, d \, \sin{\theta}}{\lambda} ##

dsinθ=mλ

If the central maximum has m=0, then sinθ=0

But if m=0, then Φ=0, and thus I=0 ...

Power density at the central maximum = 0

Secondary maximum m=1, Φ=2π

Intensity for secondary maximum is then:

I = I0 [ sin2(Nπ) / sin2(π) ]

I'm not sure what I'm missing here...
See post 4. See also post 7.
 
  • #12
Charles Link
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Additional note: This formula is kind of a tricky one because the denominator does go to zero at the primary maxima, which is an important point where the intensity value is needed. Because of that, it is kind of a clumsy formula that you need to know how to work with it. In places where the denominator goes to zero, you take the limit as described in post 4. ## \\ ## The formula is actually quite exact, and as clumsy as it is for the first couple of times until you learn how to use it, it is in widespread use because it is very useful and quite exact.
 
  • #13
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Ok, so taking the limit, letting Φ/2 = mπ+δ

I = I0 [ sin2 (N(mπ+δ)) / sin2(mπ+δ) ]

which can also be expressed as:

I = I0 [ sin2 (Nmπ+Nδ)) / sin2(mπ+δ) ]
 
  • #14
Charles Link
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Good. ## \\ ## Now observe: ## \\ ## ## \sin(Nm \pi+N \delta)=sin(Nm \pi)cos(N \delta)+cos(Nm \pi)sin(N \delta) = \pm sin(N \delta) ##. For the denominator you get similarly ## \pm \sin(\delta) ##. Thereby the ratio is ## \frac{sin^2(N \delta)}{sin^2(\delta)}=N^2 ## as ## \delta \rightarrow 0 ##. ## \\ ## And yes, it is clumsy, but for this formula, that is how it gets evaluated. The ones doing optics have gotten used to this one=the first time you see it, it can be a difficult one to utilize.
 
  • #15
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With that equation mπ = 0, so sin(0) = 0, and cos(0) = 1, I can see how most of the equation looks, but how do you get the ± sign at the end?
 
  • #16
Charles Link
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With that equation mπ = 0, so sin(0) = 0, and cos(0) = 1, I can see how most of the equation looks, but how do you get the ± sign at the end?
## cos(Nm \pi)=\pm 1 ## and ## cos(m \pi)=\pm 1 ##. ## \\ ## Note: ## m ## can be equal to ## 0 ##, but, in general, ## m ## is any integer. ## sin(Nm \pi)=0 ## and ## sin(m \pi)=0 ## for all integer ## m ##. ## \\ ## When ## m=0 ##, that is the central maximum for the diffraction grating, but there can be numerous primary maxima (integer ## m ##) all having intensity ## I=N^2 I_o ##. ## \\ ## There is a maximum value of ## m ## because for ## m \lambda=d \sin{\theta} ##, ## |\sin{\theta}|<1 ##.
 
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  • #17
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So now for the second maxima. The second maxima will occur when NΦ/2 = Nmπ ± π/2, and Φ/2 = mπ ± π/2?
 
  • #18
Charles Link
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So now for the second maxima. The second maxima will occur when NΦ/2 = Nmπ ± π/2, and Φ/2 = mπ ± π/2?
It's called "secondary". Almost correct. You are finding the secondary maximum nearest the ## m ##th primary maximum. The numerator has the value 1. The denominator needs to get expanded. This one, you won't need to take a limit (correct that=see below). The limit is only necessary for the primary maxima. The denominator will be non-zero for this one. ## \\ ## For the denominator, you can take the limit of large ## N ##. And for the denominator, you should have ## \frac{\Phi}{2}=m \pi+\pm \frac{\pi}{2 N } ##. ## \\ ## Once you compute the intensity ## I ##, you can leave it as ##I=I_o \frac{1}{\sin^2(\frac{\pi}{2N} )} ##, but its useful to compute the large ## N ## case as ##\frac{ 4N^2}{\pi^2} ##. (Edit: Made a mistake here: See post 29) ## \\ ## Additional note: It is actually easier to find the places wh.ere the numerator is zero for this formula, and where the denominator is not zero. Between any two zeros, there is a secondary maximum. (The location is approximately where the numerator is 1, but not precisely). For this problem though, we are just interested in the secondary maximum that is right next to the primary one. There are ## N-2 ## secondary maxima between any two primary maxima, because there are ## N-1 ## zeros between any two primary maxima. There are ## N-1 ## places between where the numerator goes to zero, but not the denominator.
 
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  • #19
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sin(Nπ/2) = ±1 or 0, depending on if N is an even or odd number

cos(Nπ/2) = ±1 or 0, depending on if N is an even or odd number

For the numerator:

sin(N(mπ + π/2)) = sin(Nmπ)cos(Nπ/2)+cos(Nmπ)sin(Nπ/2)

I'm not really sure how to simplify this...

For the denominator:

sin(mπ+π/2) = sin(mπ)cos(π/2)+cos(mπ)sin(π/2)

sin(π/2) = 1

cos(π/2) = 0

∴ sin(mπ+π/2) = cos(mπ)

You said numerator = 1, so I'm guessing mπ=1

If that's the case then numerator is:

sin(N(mπ + π/2)) = sin(N)cos(Nπ/2)+cos(N)sin(Nπ/2)
 
  • #20
Charles Link
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Numerator is ## \frac{N \Phi}{2}=Nm \pi \pm\frac{\pi}{2} ## for the secondary maximum next to the ## m ##th primary maximum. The sine of that is simply ## \pm 1 ##. For the denominator, see post 18. ## \\ ## Please also see post 7 again. ## \\ ## Post 19 is incorrect. Please look closely at ## \frac{N \Phi}{2} ## and ## \frac{\Phi}{2} ##. ## \\ ## See post 18. ## \frac{\Phi}{2}=m \pi \pm \frac{\pi}{2N} ##. (Comes from ## \frac{N \Phi}{2}=Nm \pi \pm \frac{\pi}{2} ## ).
 
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  • #21
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For the denominator:

sin(mπ+π/2N) = sin(mπ)cos(π/2N)+cos(mπ)sin(π/2N)

I'm not sure how this simplifies. I'm assuming m=1, therefore mπ=π, which would make it:

sin(mπ+π/2N) = sin(π/2N)
 
  • #22
Charles Link
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For the denominator:

sin(mπ+π/2N) = sin(mπ)cos(π/2N)+cos(mπ)sin(π/2N)

I'm not sure how this simplifies. I'm assuming m=1, therefore mπ=π, which would make it:

sin(mπ+π/2N) = sin(π/2N)
## sin(m \pi)=0 ##. No reason that ## m=1 ##. This works for all ## m ##. ## \\ ## And yes, your last line is correct. (It could use a ## \pm ##, but otherwise is correct).
 
  • #23
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So the Power density of the secondary maxima:

I = I0 / sin2(π/2N)

Power density for primary maxima:

I = I0N2
 
  • #24
Charles Link
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So the Power density of the secondary maxima:

I = I0 / sin2(π/2N)

Power density for primary maxima:

I = I0N2
Very good !! :smile::smile: ## \\ ## See also post 18. The ratio will simplify for large ## N ##. I'm not quite sure whether the large ## N ## case is being asked for in the original problem statement.
 
  • #25
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I think the question just asks for the ratio of the two, so:

I0N2 / [ I0 / sin2(π/2N) ]
 

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