Question

(a) Consider an arrangement of $N$ slits with a distance $d$ between adjacent slits. The slits emit coherently and in phase at wavelength $\lambda$. Show that at a time $t$, the electric field at a distant point $P$ is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where $E_0$ is the amplitude at $P$ of the electric field due to an individual slit, $\phi = (2\pi d$ sin $\theta)/\lambda$, $\theta$ is the angle of the rays reaching $P$ (as measured from the perpendicular bisector of the slit arrangement), and $R$ is the distance from $P$ to the most distant slit. In this problem, assume that $R$ is much larger than $d$. (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $e^{iz} = cos z + i \space sin \space z$, where $i = \sqrt{-1}$. In this expression, cos $z$ is the $real$ $part$ of the complex number $e^{iz}$, and sin $z$ is its $imaginary$ $part$. Show that the electric field $E_P(t)$ is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that $e^Ae^B = e^{(A+B)}$ and $(e^A)^n = e^{nA}$, show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship $e^{iz}$ = cos $z$ + $i$ sin $z$, show that the (real) electric field at point $P$ is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at $P$. (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle $\theta$ is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where $I_0$ is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case $N$ = 2. It will help to recall that sin 2$A$ = 2 sin $A$ cos $A$. Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. ($Hint$: Is I0 defined in the same way in both expressions?)

Name: Problem 66, Chapter 36: University Physics with Modern Physics
Uploaded: 2020-12-30T04:13:08-08:00
Duration: 18 min 31 s
Channel: Sheh Lit Chang

   (a) Consider an arrangement of $N$ slits with a distance $d$ between adjacent slits. The slits emit coherently and in phase at wavelength $\lambda$. Show that at a time $t$, the electric field at a distant point $P$ is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where $E_0$ is the amplitude at $P$ of the electric field due to an individual slit, $\phi = (2\pi d$ sin $\theta)/\lambda$, $\theta$ is the angle of the rays reaching $P$ (as measured from the perpendicular bisector of the slit arrangement), and $R$ is the distance from $P$ to the most distant slit. In this problem, assume that $R$ is much larger than $d$. (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $e^{iz} = cos z + i \space sin \space z$, where $i = \sqrt{-1}$. In this expression, cos $z$ is the $real$ $part$ of the complex number $e^{iz}$, and sin $z$ is its $imaginary$ $part$. Show that the electric field $E_P(t)$ is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that $e^Ae^B = e^{(A+B)}$ and $(e^A)^n = e^{nA}$, show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship $e^{iz}$ = cos $z$ + $i$ sin $z$, show that the (real) electric field at point $P$ is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at $P$. (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle $\theta$ is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where $I_0$ is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case $N$ = 2. It will help to recall that sin 2$A$ = 2 sin $A$ cos $A$. Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. ($Hint$: Is I0 defined in the same way in both expressions?)

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 36, Problem 66

Instant Answer

Solved by Expert Sheh Lit Chang

12/30/2020

Step 1

The electric field due to each slit can be represented as $E_0 \cos (kR - \omega t + n\phi)$, where $E_0$ is the amplitude at P of the electric field due to an individual slit, $\phi = \frac{2\pi d \sin \theta}{\lambda}$, $\theta$ is the angle of the rays reaching Show more…

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(a) Consider an arrangement of $N$ slits with a distance $d$ between adjacent slits. The slits emit coherently and in phase at wavelength $\lambda$. Show that at a time $t$, the electric field at a distant point $P$ is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where $E_0$ is the amplitude at $P$ of the electric field due to an individual slit, $\phi = (2\pi d$ sin $\theta)/\lambda$, $\theta$ is the angle of the rays reaching $P$ (as measured from the perpendicular bisector of the slit arrangement), and $R$ is the distance from $P$ to the most distant slit. In this problem, assume that $R$ is much larger than $d$. (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $e^{iz} = cos z + i \space sin \space z$, where $i = \sqrt{-1}$. In this expression, cos $z$ is the $real$ $part$ of the complex number $e^{iz}$, and sin $z$ is its $imaginary$ $part$. Show that the electric field $E_P(t)$ is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that $e^Ae^B = e^{(A+B)}$ and $(e^A)^n = e^{nA}$, show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship $e^{iz}$ = cos $z$ + $i$ sin $z$, show that the (real) electric field at point $P$ is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at $P$. (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle $\theta$ is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where $I_0$ is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case $N$ = 2. It will help to recall that sin 2$A$ = 2 sin $A$ cos $A$. Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. ($Hint$: Is I0 defined in the same way in both expressions?)

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Transcript

00:01 We want to find the intensity formula for diffraction using from n -slits.

00:13 So this is the diagram that we are going to draw.

00:19 And then we want to, we are kind of guided.

00:22 So there are five parts in this question.

00:27 So in part a, we want to find a.

00:31 Electric field at point p where point p is a point on the screen okay yeah and it is very far from the far from the slate and then the slate separation is d okay so uh you are going to write down the electric fuels at point p due to from different slits okay point p on the electric field from different slits k r okay so i label one to be at the bottom and go up to n and e1 okay we'll write it right as e .0 cosine k times big r minus omega t okay and then e2 to be e0 cosine kr minus omega t plus five okay and our five is the it's a phase difference due to path difference okay okay so we have this relationship by divided by two pi equals to lambda over uh no equals to d sine data over lambda okay so phi is equal to two pi d sine theta over lambda okay so this is our phase difference and then you can do e3, e0, cosine, ar minus omega t plus 2 ,5.

02:33 And then you keep doing this.

02:36 And then you get the electric field due to the n's sleep.

02:41 And you get e .0 cosine, ar minus omega -t plus and minus 1 ,5.

02:50 And then the electric field at point p at a time, at an instance.

02:57 Will be e1 plus e2 plus e3 plus up until en.

03:07 So it will be e not cosine, kr minus omega t plus e not cosine kr minus omega t plus five plus e not cosine kr minus omega t plus five plus e not cosine kr minus omega t plus omega, t plus two five and so on until you reach the last term okay then is the n n cosine they are minus omega t plus and minus one five okay yeah and shown okay so this is how you obtain the electric field point p okay by just summing up you want e 2 e 2 e3 until en okay so just summing up out all the contributions by each from each sleep okay all right okay then in prime b okay we are asked to show that this thing okay the summation okay the real part of this thing is equal to e pt okay you know to point p so so the first thing we'll do is we'll use the euler formula okay so eg to the i data right um right into the i data equals to cosine data plus i sign data okay so you can use this relationship and you can write uh there's a e not okay e not cosine kr minus omega t plus n5 uh plus i sign kr minus omega t plus n5 uh plus i sign k r minus omega t plus n5 okay and if you take the real part of the summation okay then you are just going to extract the cosine term okay because that's the real part okay then if you expand the summation okay so you start from the n equals to zero and then you do n equals to one okay and n equals to two and then continue on you get to the last term and you realize that this is just ept okay and shown all right so that's how you show that the real part of the complex exponential some of the complex exponential gives you the ept okay so why we want to do this is that it is easier to do the summation in complex exponentials and then later where we want to find the ept, then we just take the real part of that sum.

07:12 Okay, so that's a rationale.

07:16 Then we go on to part c.

07:19 In prat c, you want to show that the summation of the complex exponential.

07:26 It can be written, can be simplified into the given expression.

07:31 So i'm just going to proceed on to show how you can obtain that.

07:36 Okay.

07:40 Okay, so this is the given.

07:43 Summation, okay.

07:45 The first thing is i'm going to pull out the constant term, the terms that does not involve n, small n.

07:56 So e0, e.

07:57 E to the i .kr minus omega t, that does not contain n, so you can pull out from the summation and what you are left with is summation from n equals to 0 to n minus 1, e to the i.

08:11 By okay so this thing is a geometric sum okay so you can write we can use the formula for the geometric sum okay okay so you have the last term minus the first term divide by the ratio minus the first term so then you can put this, put this thing into here, okay, so if you know, e to the i, kr minus omega t, e to the i m5 minus 1, divide by, e to the i5, minus 1.

09:26 Okay, and then we can pull out some extra, some term, so on the numerator we are going to put out e -i -n -5 devoured by 2 and then we have e to the i -n -5 diva -2 minus e -2 minus i n5 divided by 2.

09:50 Okay and then the denominator we are going to pull out e - to the i -5 devout by 2 so we have e to the i -5 devout by 2 minus e to the minus i -5 over 2.

10:03 Okay, and then we are going to absorb this term into here.

10:17 Okay, so we have e0 e, i, kr minus omega t, plus n minus 1, diva by 2, 5, e, e, dita, i, and 5, i .m .5, e, dita, i .m .5, 0 by 2, minus, e, y, dada, i and phi over 2 divide by e to the i5 over 2 minus e to the minus i5 over 2 okay all right so this is what we managed to get okay and then we want to find a real part okay so so we want to find electric field at point p okay is equal to the real parts of the summation it is also equal to the real part of the term above your part, okay.

12:20 So when it comes to taking the real part, this term will get the cosine...

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