Reflection and Refraction of a Plane Wave at an Air–Water Interface: 04-BS-9, December 2018, Q6

Question 6: Reflection and Refraction of a Plane Wave at an Air–Water Interface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Frequency $f$	$1\times10^{10}$ Hz
Angle of incidence $\theta_1$ (from normal)	$45^\circ$, in air
Relative permittivity of water $\varepsilon_r$	$81$ (assume $\mu_r\approx1$)

Find. The direction and wavelength of (a) the reflected wave and (b) the penetrating (transmitted/refracted) wave.

Incident ray at 45° from the normal; the reflected ray leaves at the same 45° back into air, while the transmitted ray bends sharply toward the normal (4.51°) entering the optically denser water.

Approach. The law of reflection gives the reflected ray directly (same medium, same angle); Snell's law with $n_2=\sqrt{\varepsilon_r}$ gives the refracted ray's angle; wavelength in each medium follows from $\lambda=v/f=c/(nf)$.

Reflected wave. It never leaves air, so its angle equals the angle of incidence (on the opposite side of the normal) and its wavelength is unchanged from the incident wave: $$\theta_{\text{refl}}=45^\circ,\qquad \lambda_1=\frac{c}{f}=\frac{3\times10^8}{1\times10^{10}}$$ $$\boxed{\lambda_1=0.0300\ \text{m}=3.00\ \text{cm}}$$
Refracted (penetrating) wave direction. Water's refractive index $n_2=\sqrt{\varepsilon_r}=\sqrt{81}=9$. Snell's law: $$\sin\theta_2=\frac{\sin\theta_1}{n_2}=\frac{\sin45^\circ}{9}=0.0786$$ $$\boxed{\theta_2=\sin^{-1}(0.0786)=4.51^\circ}$$ from the normal — bent sharply TOWARD the normal, since water is the optically denser medium.
Refracted wavelength. Using $v_2=c/n_2$: $$\lambda_2=\frac{v_2}{f}=\frac{\lambda_1}{n_2}=\frac{0.0300}{9}$$ $$\boxed{\lambda_2=3.33\times10^{-3}\ \text{m}=3.33\ \text{mm}}$$

Quantity	Result
Reflected ray angle	$45^\circ$ (from normal, in air)
Reflected wavelength $\lambda_1$	$3.00$ cm
Refracted ray angle	$4.51^\circ$ (from normal, in water)
Refracted wavelength $\lambda_2$	$3.33$ mm

Concept & Theory

Topic. Reflection and refraction (Snell's law) of a plane EM wave at a dielectric interface.

Key relations. $\theta_{\text{refl}}=\theta_{\text{inc}}$; $n_1\sin\theta_1=n_2\sin\theta_2$ with $n=\sqrt{\varepsilon_r}$ (nonmagnetic media); $\lambda=v/f=c/(nf)$.

Why this works. The tangential component of the wavevector must match across the boundary at every point and instant (continuity of phase along the interface) — this is exactly Snell's law. The wave slows and its wavelength shrinks by the same factor $n$ on entering the denser medium, while the frequency (fixed by the source) is unchanged.

Common pitfall. Using $n=\varepsilon_r$ instead of $n=\sqrt{\varepsilon_r}$, or computing the transmitted wavelength from the source frequency and the WRONG medium's propagation speed.

Source. Sadiku, Elements of Electromagnetics, Ch. 10 (EM wave propagation); Young & Freedman, University Physics, Ch. 33 (reflection/refraction, Snell's law).

Question 6 of 8: Reflection and Refraction of a Plane Wave at an Air–Water Interface

Question 6: Reflection and Refraction of a Plane Wave at an Air–Water Interface (20 marks)