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Some minor points: First, the units of $\lambda^2$ will be $(\mu m)^2$. Second it seems you are asking for $a, b$ so that you can determine $n$? In fact, what is more often done is to plot $n$ versus $1/\lambda^2$ to determine $a, b$. At the moment you can't do anything (unless you follow Frobenius and look up $a, b$, however this means you know the ...


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Wavelength is used as a convenience. It's much easier to imagine a photon with a 500 nm wavelength than to comprehend a photon oscillating 600 trillion times per second. But in reality that's all it is is a photon moving at the speed of light and oscillating 600 trillion times per second as it goes along. The photon completes one cycle every 500 nm. Many on ...


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The speed of sound depends on the medium in which it travels. In air it can be affected by ambient temp, relative humidity , as well as atmospheric pressure. The speed usual quoted at a standard temp and pressure. http://byjus.com/physics/speed-of-sound-propagation/


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Coherent sources- Two sources of light are said to be coherent if the waves emitted from them have the same frequency and are 'phase-linked'; that is, they have a zero or constant phase difference. The calculation yielded 90.pi ; We know that 2.pi denotes a phase change of zero as the waves will come back to initial phase relation of the coherent ...


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Coherency has to be over a period of time. Instantaneously any two waves (or signals) at the same freq will have at at one time or point a phase difference. If their freqs stay exactly what they had at one time, they will stay coherent with each other, i.e. The phase differences won't change over the same period of time for the two. Over a period of time the ...


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When working with waves the wavelength and frequency (pitch) are inversely related. Sound waves have the relation frequency times wavelength equal the speed of sound. Wind instruments are using the longitudinal dimension of air in the instrument as a medium like a "string". In both the string and wind instruments, shorter wavelengths in the excited medium ...


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Wind instruments work by setting up sanding waves in the air column inside them. Shorter instruments have shorter air columns and thus standing waves with shorter wavelengths resulting in higher pitches.


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Your error is writing this expression $$A_{\text{ res}}\cos(kx-\omega t) = A_1\cos(kx-\omega t) + A_2\cos(kx-\omega t -\phi)$$ It should be $$A_{\text{ res}}\cos(kx-\omega t-\psi) = A_1\cos(kx-\omega t) + A_2\cos(kx-\omega t -\phi)$$ Note the resultant is not in phase with $A_1\cos(kx-\omega t)$ I think that a simple way of doing the addition is to draw ...


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What is wrong is to assume that the sum will read $A_{res}\cos(\omega t - kx)$ while in fact the sum should read in general $A_{res}\cos(\omega t - k x + \phi_{res} )$. The strategy consists in starting from the sought solution $y_{res}= A_{res}\cos(\omega t - k x + \phi_{res})$ and expand the cosine function $y_{res} = A_{res}[\cos(\omega t - k x)\cos ...


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Whether the amount of diffraction is 'negligible' depends on how you define this criterion. The first order minimum in the diffraction pattern from a single slit occurs where $\sin\theta = \lambda/d$ where $d$ is slit width, $\theta$ is diffraction angle and $\lambda$ is wavelength. If $d = \lambda$ the central lobe of the diffraction pattern will spread ...



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