The electric fields of two EM waves in vacuum are both described by: $$\mathbf{E} = E_0 \sin(kx-\omega t)\hat{y}$$ The "wave number" $k$ of wave 1 is larger than that of wave 2, $k_1 > k_2$. Which wave has the larger frequency $f$? 1. Wave 1 2. Wave 2 3. impossible to tell Note: * Correct Answer: A * Same speed and thus wavelength of 1 is smaller, so frequency is higher

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For a wave on a 1d string that hits a boundary between 2 strings of different material we get, $$\widetilde{f}(z<0) = \widetilde{A}_I e^{i(k_1)z-\omega t} + \widetilde{A}_Re^{i(-k_1z-\omega t)}$$ $$\widetilde{f}(z>0) = \widetilde{A}_T e^{i(k_2)z-\omega t}$$ where continuity (BCs) give, $$\widetilde{A}_R = \left(\dfrac{k_1-k_2}{k_1+k_2}\right)\widetilde{A}_I$$ $$\widetilde{A}_T = \left(\dfrac{2k_1}{k_1+k_2}\right)\widetilde{A}_I$$ Is the transmitted wave in phase with the incident wave? A) Yes, always B) No, never C) Depends Note: * Correct answer: A

For a wave on a 1d string that hits a boundary between 2 strings of different material we get, $$\widetilde{f}(z<0) = \widetilde{A}_I e^{i(k_1)z-\omega t} + \widetilde{A}_Re^{i(-k_1z-\omega t)}$$ $$\widetilde{f}(z>0) = \widetilde{A}_T e^{i(k_2)z-\omega t}$$ where continuity (BCs) give, $$\widetilde{A}_R = \left(\dfrac{k_1-k_2}{k_1+k_2}\right)\widetilde{A}_I$$ $$\widetilde{A}_T = \left(\dfrac{2k_1}{k_1+k_2}\right)\widetilde{A}_I$$ Is the reflected wave in phase with the incident wave? A) Yes, always B) No, never C) Depends Note: * Correct answer: C * Can be 180 out of phase

In matter we have, $$\nabla \cdot \mathbf{D} = \rho_f \qquad \nabla \cdot \mathbf{B} = 0$$ $$\nabla \times \mathbf{E} = -\dfrac{\partial \mathbf{B}}{\partial t} \qquad \nabla \times \mathbf{H} = \mathbf{J}_f + \dfrac{\partial \mathbf{D}}{\partial t}$$ with $$\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} \qquad \mathbf{H} = \mathbf{B}/\mu_0 - \mathbf{M}$$ If there are no free charges or current, is $\nabla \cdot \mathbf{E} = 0$? 1. Yes, always 2. Yes, under certain conditions (what are they?) 3. No, in general this will not be true 4. ?? Note: * Correct answer: B

In linear dielectrics, $\mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P} = \varepsilon \mathbf{E}.$ In a linear dielectric is $\varepsilon > \varepsilon_0$? 1. Yes, always 2. No, never 3. Sometimes, it depends on the details of the dielectric. Note: * Correct answer: A

In a non-magnetic, linear dielectric, $$v = \dfrac{1}{\sqrt{\mu \varepsilon}} = \dfrac{1}{\sqrt{\mu \varepsilon_r \varepsilon_0}} = \dfrac{c}{\sqrt{\varepsilon_r}}$$ How does $v$ compare to $c$? 1. $v>c$ always 2. $v<c$ always 3. It depends Note: * Correct Answer: B