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Question about the derivation of the intensity formula of a diffraction grating

Simulating the Interference Pattern of Fraunhofer Diffraction by a Single SlitDiffraction Grating Spectrometry QuestionCovering centeremost slit of a N slit diffraction grating - what happens?Resolving power of a diffraction grating?Optics Diffraction Grating PlotTotal number of primary maxima in diffraction gratingDeriving formula for effect of slit width and multiplicity for multi-slit diffraction patternIntensity at $theta = 0$ in slit diffractionBasic, intuitive explanation for a diffraction gratingIntensity function of diffraction confusion

4

$begingroup$

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:

$$I = I_0
left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)^2
left( fracsin(kasintheta)kasintheta right)^2
$$

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

share|cite|improve this question

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

$endgroup$

add a comment | 

4

$begingroup$

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:

$$I = I_0
left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)^2
left( fracsin(kasintheta)kasintheta right)^2
$$

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

share|cite|improve this question

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

$endgroup$

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:

$$I = I_0
left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)^2
left( fracsin(kasintheta)kasintheta right)^2
$$

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

share|cite|improve this question

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

$endgroup$

share|cite|improve this question

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

share|cite|improve this question

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

edited Apr 3 at 16:53

Thomas Fritsch

1,548515

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

asked Apr 3 at 14:45

A. PavlenkoA. Pavlenko

414

1 Answer
1

active

oldest

votes

5

$begingroup$

(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
$$
beginalign
E(theta)
&= E_0 sum_n=-N^+N int_-a^+a e^ik(nd+x)sintheta textdx \
&= E_0 left( sum_n=-N^+N e^ikndsinthetaright)
left( int_-a^+a e^ikxsintheta textdx right) \
&= E_0 left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)
left( 2afracsin(kasintheta)kasintheta right)
endalign
$$

And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$

share|cite|improve this answer

edited Apr 3 at 20:50

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515

$endgroup$

add a comment | 

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5

$begingroup$

(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
$$
beginalign
E(theta)
&= E_0 sum_n=-N^+N int_-a^+a e^ik(nd+x)sintheta textdx \
&= E_0 left( sum_n=-N^+N e^ikndsinthetaright)
left( int_-a^+a e^ikxsintheta textdx right) \
&= E_0 left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)
left( 2afracsin(kasintheta)kasintheta right)
endalign
$$

And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$

share|cite|improve this answer

edited Apr 3 at 20:50

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515

$endgroup$

add a comment | 

5

$begingroup$

(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
$$
beginalign
E(theta)
&= E_0 sum_n=-N^+N int_-a^+a e^ik(nd+x)sintheta textdx \
&= E_0 left( sum_n=-N^+N e^ikndsinthetaright)
left( int_-a^+a e^ikxsintheta textdx right) \
&= E_0 left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)
left( 2afracsin(kasintheta)kasintheta right)
endalign
$$

And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$

share|cite|improve this answer

edited Apr 3 at 20:50

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515

$endgroup$

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
$$
beginalign
E(theta)
&= E_0 sum_n=-N^+N int_-a^+a e^ik(nd+x)sintheta textdx \
&= E_0 left( sum_n=-N^+N e^ikndsinthetaright)
left( int_-a^+a e^ikxsintheta textdx right) \
&= E_0 left( fracsin((N+frac12)kdsintheta)sin(frac12kdsintheta) right)
left( 2afracsin(kasintheta)kasintheta right)
endalign
$$

And finally you get the intensity by taking the absolute square of the amplitude
$$I(theta) = |E(theta)|^2$$

share|cite|improve this answer

edited Apr 3 at 20:50

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515

$endgroup$

share|cite|improve this answer

edited Apr 3 at 20:50

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515

answered Apr 3 at 15:43

Thomas FritschThomas Fritsch

1,548515