5 Used the word "amplitude" in a few places where I shouldn't have
source | link

At the screen, the amplitudewave due to the right source is $A\sin\big(kD-\omega t\big)$. The amplitudewave due to the left source is $A\sin\big( k(D+d\sin\theta)-\omega t \big)$. Therefore, our amplitudetotal wave is

Ooookay. One more thing. If we back up, we need to clarify that $k$ is called the wave number. It is related to wavelength by $k = 2\pi/\lambda$. With this, we know $\Delta\phi = kd\sin\theta = 2\pi d\sin\theta/\lambda$. Therefore, the amplitudewave is $2A\cos\left(\frac{\pi d\sin\theta}{\lambda}\right)$. The intensity is given by the square of the amplitude, so $$ I(\theta) = 4A^{2}\cos^{2}\left(\frac{\pi d\sin\theta}{\lambda}\right). $$

We have a single slit of width $a$. The strategy is to split the wave into $N$ waves in the spirit of Hyugens. The wave sources are equally spaced out by $\Delta s = a/N$. Each wave source contributes $1/N$ of the original amplitudewave.

We are now going to use a trick. The trick is to take advantage of Euler's identity $e^{iu} = \cos(u)+i\sin(u)$. This will seem really cheap, but we will simply replace all of the $\cos u$ terms by $e^{iu}$, and we will understand that we're dealing only with the real part when necessary.

Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals $$ \int_{-\frac{d}{2}+\frac{a}{2}}^{-\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}+\frac{a}{2}}^{\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$$$ \int_{-\frac{d}{2}-\frac{a}{2}}^{-\frac{d}{2}+\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}-\frac{a}{2}}^{\frac{d}{2}+\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$ We need to evaluate the integrals, and do serious trig acrobatics. Evaluating and simplifying a few things gives $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$ We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$ As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to $$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$ and this is $$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$ By squaring the amplitude, we obtain $$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

At the screen, the amplitude due to the right source is $A\sin\big(kD-\omega t\big)$. The amplitude due to the left source is $A\sin\big( k(D+d\sin\theta)-\omega t \big)$. Therefore, our amplitude is

Ooookay. One more thing. If we back up, we need to clarify that $k$ is called the wave number. It is related to wavelength by $k = 2\pi/\lambda$. With this, we know $\Delta\phi = kd\sin\theta = 2\pi d\sin\theta/\lambda$. Therefore, the amplitude is $2A\cos\left(\frac{\pi d\sin\theta}{\lambda}\right)$. The intensity is given by the square of the amplitude, so $$ I(\theta) = 4A^{2}\cos^{2}\left(\frac{\pi d\sin\theta}{\lambda}\right). $$

We have a single slit of width $a$. The strategy is to split the wave into $N$ waves in the spirit of Hyugens. The wave sources are equally spaced out by $\Delta s = a/N$. Each wave source contributes $1/N$ of the original amplitude.

We are now going to use a trick. The trick is to take advantage of Euler's identity $e^{iu} = \cos(u)+i\sin(u)$. This will seem really cheap, but we will simply replace all of the $\cos u$ terms by $e^{iu}$, and we will understand that we're dealing only with the real part when necessary.

Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals $$ \int_{-\frac{d}{2}+\frac{a}{2}}^{-\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}+\frac{a}{2}}^{\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$ We need to evaluate the integrals, and do serious trig acrobatics. Evaluating and simplifying a few things gives $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$ We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$ As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to $$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$ and this is $$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$ By squaring the amplitude, we obtain $$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

At the screen, the wave due to the right source is $A\sin\big(kD-\omega t\big)$. The wave due to the left source is $A\sin\big( k(D+d\sin\theta)-\omega t \big)$. Therefore, our total wave is

Ooookay. One more thing. If we back up, we need to clarify that $k$ is called the wave number. It is related to wavelength by $k = 2\pi/\lambda$. With this, we know $\Delta\phi = kd\sin\theta = 2\pi d\sin\theta/\lambda$. Therefore, the wave is $2A\cos\left(\frac{\pi d\sin\theta}{\lambda}\right)$. The intensity is given by the square of the amplitude, so $$ I(\theta) = 4A^{2}\cos^{2}\left(\frac{\pi d\sin\theta}{\lambda}\right). $$

We have a single slit of width $a$. The strategy is to split the wave into $N$ waves in the spirit of Hyugens. The wave sources are equally spaced out by $\Delta s = a/N$. Each wave source contributes $1/N$ of the original wave.

We are now going to use a trick. The trick is to take advantage of Euler's identity $e^{iu} = \cos(u)+i\sin(u)$. This will seem cheap, but we will simply replace all of the $\cos u$ terms by $e^{iu}$, and we will understand that we're dealing only with the real part when necessary.

Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals $$ \int_{-\frac{d}{2}-\frac{a}{2}}^{-\frac{d}{2}+\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}-\frac{a}{2}}^{\frac{d}{2}+\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$ We need to evaluate the integrals, and do serious trig acrobatics. Evaluating and simplifying a few things gives $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$ We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$ As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to $$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$ and this is $$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$ By squaring the amplitude, we obtain $$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

4 Added the derivations.
source | link

We're dealing with 2D waves where they spread out in a circle from each source, but along each line we essentially have a 1D wave. In the picture above, if $x$ is the distance along one of the arrows, the wave is given by $A\sin(kx-\omega t)$. If the right arrow has total distance $x=D$, the left arrow has total distance $x=D+d\sin\theta$ (I'm getting a little confused myself, but in any case the ideaall that matters is that if you draw a really long triangle, long triangle, I would bet that the two longest sides would differ by about $d\sin\theta$).

The key to this is that the "real part" is additive, so $\text{Re}\; (e^{iu}+e^{iu'}) = \text{Re}\; (e^{iu}) + \text{Re}\; (e^{iu'})$. Note: this(this is not so simple if we are multiplying complex numbers though). This only works because we are adding things. Also, what we're doing respects integration, so $\int_{a}^{b} \text{Re}\;e^{iu}\;du = \text{Re}\;\int_{a}^{b}e^{iu}\;du$. The magic to this trick is that it makes all of the trigonometry unbelievably easy.

Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals $$ \int_{-\frac{d}{2}+\frac{a}{2}}^{-\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}+\frac{a}{2}}^{\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$ We need to evaluate the integrals, and do some serious trig acrobatics. Evaluating and simplifying a few things gives $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$ We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$ As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to $$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$ and this is $$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$ By squaring the amplitude, we obtain $$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

We're dealing with 2D waves where they spread out in a circle from each source, but along each line we essentially have a 1D wave. In the picture above, if $x$ is the distance along one of the arrows, the wave is given by $A\sin(kx-\omega t)$. If the right arrow has total distance $x=D$, the left arrow has total distance $x=D+d\sin\theta$ (I'm getting a little confused myself, but in any case the idea is that if you draw a really long triangle, I would bet that the two longest sides would differ by about $d\sin\theta$).

The key to this is that the "real part" is additive, so $\text{Re}\; (e^{iu}+e^{iu'}) = \text{Re}\; (e^{iu}) + \text{Re}\; (e^{iu'})$. Note: this is not so simple if we are multiplying complex numbers. This only works because we are adding things. Also, what we're doing respects integration, so $\int_{a}^{b} \text{Re}\;e^{iu}\;du = \text{Re}\;\int_{a}^{b}e^{iu}\;du$. The magic to this trick is that it makes all of the trigonometry unbelievably easy.

Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals $$ \int_{-\frac{d}{2}+\frac{a}{2}}^{-\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}+\frac{a}{2}}^{\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$ We need to evaluate the integrals, and do some serious trig acrobatics. Evaluating and simplifying a few things gives $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$ We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$ As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to $$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$ and this is $$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$ By squaring the amplitude, we obtain $$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

We're dealing with 2D waves where they spread out in a circle from each source, but along each line we essentially have a 1D wave. In the picture above, if $x$ is the distance along one of the arrows, the wave is given by $A\sin(kx-\omega t)$. If the right arrow has total distance $x=D$, the left arrow has total distance $x=D+d\sin\theta$ (all that matters is that if you draw a really long triangle, the two longest sides would differ by about $d\sin\theta$).

The key to this is that the "real part" is additive, so $\text{Re}\; (e^{iu}+e^{iu'}) = \text{Re}\; (e^{iu}) + \text{Re}\; (e^{iu'})$ (this is not so simple if we are multiplying complex numbers though). This only works because we are adding things. Also, what we're doing respects integration, so $\int_{a}^{b} \text{Re}\;e^{iu}\;du = \text{Re}\;\int_{a}^{b}e^{iu}\;du$. The magic to this trick is that it makes all of the trigonometry unbelievably easy.

Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals $$ \int_{-\frac{d}{2}+\frac{a}{2}}^{-\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}+\frac{a}{2}}^{\frac{d}{2}-\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$ We need to evaluate the integrals, and do serious trig acrobatics. Evaluating and simplifying a few things gives $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$ We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields $$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$ As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to $$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$ and this is $$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$ By squaring the amplitude, we obtain $$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

3 Added the derivations.
source | link

I will provide the derivations for the formulas below. A discussion of this and the equations are provided in the last chapter of Vibration and Waves by A.P. French.


Two Point Sources

Two Point Sources

I will provide the derivations for the formulas below. A discussion of this and the equations are provided in the last chapter of Vibration and Waves by A.P. French.


Two Point Sources

2 added 7718 characters in body
source | link
1
source | link