Dielectric Medium Properties
The electromagnetic wave travels in a dielectric medium with permittivity $\epsilon = 4\epsilon_0$ and permeability $\mu = \mu_0$. The refractive index of the medium is calculated as:
$ n_1 = \sqrt{\frac{\epsilon \mu}{\epsilon_0 \mu_0}} = \sqrt{\frac{(4\epsilon_0)(\mu_0)}{\epsilon_0 \mu_0}} = \sqrt{4} = 2 $
The medium is air, with refractive index $n_2 = 1$. The boundary is the plane $z=0$. The wave is incident from the medium ($n_1=2$) to air ($n_2=1$).
Incident Wave Analysis
The magnetic field is given by $\vec{H} = \hat{j} H_0 \cos(\omega t - kx - k\sqrt{3}z)$. The phase of the wave is $\phi = \omega t - kx - k\sqrt{3}z$. The wave vector $\vec{k}_{inc}$ is found from the gradient of the phase:
$ \vec{k}_{inc} = \nabla \phi = -\hat{i} \frac{\partial \phi}{\partial x} - \hat{k} \frac{\partial \phi}{\partial z} = -\hat{i}(k) - \hat{k}(k\sqrt{3}) $
The direction of propagation is $-\vec{k}_{inc} = \hat{i}k + \hat{k}k\sqrt{3}$. The normal to the boundary ($z=0$) is $\hat{n} = \hat{k}$.
Angle of Incidence Calculation
The angle of incidence ($\theta_i$) is the angle between the direction of propagation and the normal to the surface.
$ \cos \theta_i = \frac{(-\vec{k}_{inc}) \cdot \hat{n}}{| -\vec{k}_{inc} | |\hat{n}|} = \frac{(\hat{i}k + \hat{k}k\sqrt{3}) \cdot \hat{k}}{\sqrt{k^2 + (k\sqrt{3})^2} \cdot 1} $
$ \cos \theta_i = \frac{k\sqrt{3}}{\sqrt{k^2 + 3k^2}} = \frac{k\sqrt{3}}{\sqrt{4k^2}} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2} $
Therefore, the angle of incidence is $\theta_i = 30^\circ$.
Reflection Angle
According to the law of reflection, the angle of reflection ($\theta_r$) is equal to the angle of incidence ($\theta_i$).
$ \theta_r = \theta_i = 30^\circ $
Refraction Angle Calculation
Snell's Law relates the angles of incidence and refraction ($\theta_t$) to the refractive indices of the two media:
$ n_1 \sin \theta_i = n_2 \sin \theta_t $
Substituting the values:
$ (2) \sin(30^\circ) = (1) \sin \theta_t $
$ 2 \times \frac{1}{2} = \sin \theta_t $
$ \sin \theta_t = 1 $
Thus, the angle of refraction is $\theta_t = 90^\circ$. This indicates that the refracted wave travels along the boundary surface between the medium and air (critical angle condition).
Conclusion
The angles of reflection and refraction are $30^\circ$ and $90^\circ$, respectively.
Important Questions from Reflection and Refraction
A monochromatic plane wave is incident normally from a dielectric medium $A$ onto another dielectric medium $B$. The indices of refraction satisfy $n_A < n_B$. One-fourth of the incident energy is reflected back into medium $A$. Let $E$ be the resultant electric field due to the superposition of the incident wave and the reflected wave. Then, the ratio of the two time-averages $\langle \vec{E}^2 \rangle_{\text{min}}/\langle \vec{E}^2 \rangle_{\text{max}}$ is
- An electromagnetic wave is incident from vacuum normally on a planar surface of a non-magnetic medium. If the amplitude of the electric field of the incident wave is $E_0$ and that of the transmitted wave is $2E_0/3$, then neglecting any loss, the refractive index of the medium is
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- A leaf appears green in daylight. If this leaf were observed in red light, what colour would it appear to have?
A beam of unpolarized light in a medium with dielectric constant $\epsilon_1$ is reflected from a plane interface formed with another medium of dielectric constant $\epsilon_2 = 3\epsilon_1$. The two media have identical magnetic permeability. If the angle of incidence is $60^\circ$, then the reflected light
