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4

The result $\omega^2=\frac{c_1+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}$ leads to two real solutions for $\omega^2$, since $ -1 \lt \cos ka \lt 1$ and the square root lies between $|c_1 - c_2|$ (for $\cos ka = -1$) and $c_1 + c_2$ (for $\cos ka = +1$), so that $\omega^2$ is always positive. The frequency $\omega$ is taken as a positive ...


4

Because by expanding the sinus term into a taylor expansion, you get $\sin(x)\approx x - \frac{x^3}{6} +\cdots$ So, for small values of k you are allowed to take just the linear term.


2

You're simply meant to solve your equation $ n_1sin(\theta_1) = 1$ for $\theta_1$ at one wavelength of light. $n_1$ is (assumedly) known at that wavelength, and $\theta_1$ is the unknown critical angle. I don't think I've ever seen an undergraduate physics question - at least not about Snell's law, that asks a student to work with wavelength varying ...



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