GATE 2018 EC Q.49: solution and answer

Official answer: 1.9 to 2.1Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Compare the phase with $\omega t-\mathbf{k}\cdot \mathbf{r}$. This gives $\omega =6\sqrt{3}\pi \times 10^8\text{rad/s}$ and $\mathbf{k}=2\pi \left({\hat{a}}_x+\sqrt{2}{\hat{a}}_z\right)\text{rad/m}$. So $k_x=2\pi$, $k_z=2\sqrt{2}\pi$ and $|\mathbf{k}|=2\pi \sqrt{1+2}=2\sqrt{3}\pi \text{rad/m}$. Check against free space, where $c=1/\sqrt{{\mu }_0{\epsilon }_0}=3\times 10^8\text{m/s}$ (${\mu }_0=4\pi \times 10^{-7}\text{H/m}$, ${\epsilon }_0\approx 8.854\times 10^{-12}\text{F/m}$): $\omega /c=\frac{6\sqrt{3}\pi \times 10^8}{3\times 10^8}=2\sqrt{3}\pi$. It matches.

2. Check that the wave is transverse: ${\mathbf{E}}_0\cdot \mathbf{k}\propto (\sqrt{2})(1)+(-1)(\sqrt{2})=0$. So this is a valid uniform plane wave.

3. The interface is the plane $x=0$, so its normal is ${\hat{a}}_x$. Both $k_x$ and $k_z$ are positive, so the wave travels towards $+x$, from free space into the dielectric, as the figure shows. The plane of incidence contains $\mathbf{k}$ and the normal, so it is the $xz$-plane.

4. $\mathbf{E}$ has only $x$ and $z$ components, so it lies in the plane of incidence, and $\mathbf{H}$ lies along $\pm {\hat{a}}_y$. This is parallel (TM) polarisation.

5. Measure the angle of incidence from the normal (the $x$-axis): $\cos {\theta }_i=\frac{k_x}{|\mathbf{k}|}=\frac{1}{\sqrt{3}}$ and $\tan {\theta }_i=\frac{k_z}{k_x}=\frac{2\sqrt{2}\pi }{2\pi }=\sqrt{2}$, so ${\theta }_i={\tan }^{-1}\sqrt{2}\approx 54.74^{\circ }$. The figure is not drawn to scale, so take the angle from $\mathbf{k}$ and not from the sketch.

6. For parallel polarisation the reflection coefficient is ${\Gamma }_{\parallel }=\frac{{\eta }_2\cos {\theta }_t-{\eta }_1\cos {\theta }_i}{{\eta }_2\cos {\theta }_t+{\eta }_1\cos {\theta }_i}$. It is zero at the Brewster angle. When ${\mu }_1={\mu }_2={\mu }_0$, that angle is $\tan {\theta }_B=\sqrt{\frac{{\epsilon }_{r2}}{{\epsilon }_{r1}}}=\sqrt{{\epsilon }_r}$, because medium 1 is free space. (For perpendicular polarisation between non-magnetic media no such angle exists. Here the wave is parallel-polarised, so 'no reflected wave' means it is incident at ${\theta }_B$.)

7. Set ${\theta }_i={\theta }_B$: $\sqrt{{\epsilon }_r}=\tan {\theta }_i=\sqrt{2}$, so ${\epsilon }_r=2.00$. Check with Snell's law: $\sin {\theta }_t=\frac{\sin {\theta }_i}{\sqrt{{\epsilon }_r}}=\frac{\sqrt{2}/\sqrt{3}}{\sqrt{2}}=\frac{1}{\sqrt{3}}$. This gives ${\theta }_t\approx 35.26^{\circ }$ and ${\theta }_i+{\theta }_t=90^{\circ }$, which is exactly the Brewster condition.