Question about the derivation of the intensity formula of a diffraction grating
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In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:
$$I = I_0
left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
left( frac{sin(kasintheta)}{kasintheta} right)^2
$$
They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?
waves diffraction
$endgroup$
add a comment |
$begingroup$
In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:
$$I = I_0
left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
left( frac{sin(kasintheta)}{kasintheta} right)^2
$$
They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?
waves diffraction
$endgroup$
add a comment |
$begingroup$
In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:
$$I = I_0
left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
left( frac{sin(kasintheta)}{kasintheta} right)^2
$$
They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?
waves diffraction
$endgroup$
In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:
$$I = I_0
left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
left( frac{sin(kasintheta)}{kasintheta} right)^2
$$
They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?
waves diffraction
waves diffraction
edited 13 hours ago
Thomas Fritsch
1,546515
1,546515
asked 15 hours ago
A. PavlenkoA. Pavlenko
344
344
add a comment |
add a comment |
1 Answer
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$begingroup$
(image from Antonine education)
The light amplitude $E(theta)$ into direction $theta$ can be calculated
straight-forward by summing the contributions
- of all the slits ($n$ from $-N$ to $+N$)
- and of the parts of each individual slit ($x$ from $-a$ to $+a$)
The path difference of each contributing ray
(compared to the path length of the ray originating from the center of the grating)
is $(nd+x)sintheta$.
And hence its phase is $k(nd+x)sintheta$.
Summing these contributions you get
$$
begin{align}
E(theta)
&= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
&= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
&= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
left( 2afrac{sin(kasintheta)}{kasintheta} right)
end{align}
$$
And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
(image from Antonine education)
The light amplitude $E(theta)$ into direction $theta$ can be calculated
straight-forward by summing the contributions
- of all the slits ($n$ from $-N$ to $+N$)
- and of the parts of each individual slit ($x$ from $-a$ to $+a$)
The path difference of each contributing ray
(compared to the path length of the ray originating from the center of the grating)
is $(nd+x)sintheta$.
And hence its phase is $k(nd+x)sintheta$.
Summing these contributions you get
$$
begin{align}
E(theta)
&= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
&= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
&= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
left( 2afrac{sin(kasintheta)}{kasintheta} right)
end{align}
$$
And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$
$endgroup$
add a comment |
$begingroup$
(image from Antonine education)
The light amplitude $E(theta)$ into direction $theta$ can be calculated
straight-forward by summing the contributions
- of all the slits ($n$ from $-N$ to $+N$)
- and of the parts of each individual slit ($x$ from $-a$ to $+a$)
The path difference of each contributing ray
(compared to the path length of the ray originating from the center of the grating)
is $(nd+x)sintheta$.
And hence its phase is $k(nd+x)sintheta$.
Summing these contributions you get
$$
begin{align}
E(theta)
&= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
&= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
&= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
left( 2afrac{sin(kasintheta)}{kasintheta} right)
end{align}
$$
And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$
$endgroup$
add a comment |
$begingroup$
(image from Antonine education)
The light amplitude $E(theta)$ into direction $theta$ can be calculated
straight-forward by summing the contributions
- of all the slits ($n$ from $-N$ to $+N$)
- and of the parts of each individual slit ($x$ from $-a$ to $+a$)
The path difference of each contributing ray
(compared to the path length of the ray originating from the center of the grating)
is $(nd+x)sintheta$.
And hence its phase is $k(nd+x)sintheta$.
Summing these contributions you get
$$
begin{align}
E(theta)
&= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
&= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
&= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
left( 2afrac{sin(kasintheta)}{kasintheta} right)
end{align}
$$
And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$
$endgroup$
(image from Antonine education)
The light amplitude $E(theta)$ into direction $theta$ can be calculated
straight-forward by summing the contributions
- of all the slits ($n$ from $-N$ to $+N$)
- and of the parts of each individual slit ($x$ from $-a$ to $+a$)
The path difference of each contributing ray
(compared to the path length of the ray originating from the center of the grating)
is $(nd+x)sintheta$.
And hence its phase is $k(nd+x)sintheta$.
Summing these contributions you get
$$
begin{align}
E(theta)
&= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
&= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
&= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
left( 2afrac{sin(kasintheta)}{kasintheta} right)
end{align}
$$
And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$
edited 9 hours ago
answered 14 hours ago
Thomas FritschThomas Fritsch
1,546515
1,546515
add a comment |
add a comment |
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