5 Used the word "amplitude" in a few places where I shouldn't have edited Apr 14 '17 at 3:43 SpiralRain 61011 gold badge66 silver badges1616 bronze badges 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. edited Apr 14 '17 at 3:32 SpiralRain 61011 gold badge66 silver badges1616 bronze badges 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. edited Apr 14 '17 at 3:24 SpiralRain 61011 gold badge66 silver badges1616 bronze badges 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 edited Apr 14 '17 at 3:13 SpiralRain 61011 gold badge66 silver badges1616 bronze badges 1 answered Apr 13 '17 at 22:03 SpiralRain 61011 gold badge66 silver badges1616 bronze badges