C3-1. Wavefronts, Rays, Reflection, Refraction and Total Internal Reflection

1. Wavefronts and Rays

  • Wavefront: A line joining adjacent points on a wave that are completely in phase (e.g., all crests or all troughs).
  • Ray: An arrow showing the direction of wave motion and energy transfer.
  • Rays are always strictly perpendicular to wavefronts.
  • The distance between successive wavefronts is one wavelength $\lambda$.
ray λ wavefronts
λ 90° Wavefront at peak Wavefront at trough Ray

2. Wave Reflection

Law of Reflection

  • Occurs when a wave hits a boundary and bounces back into the original medium.
  • Angles are measured from the normal (the dashed line drawn at $90^\circ$ to the boundary).
  • Constant properties: The frequency, wavelength, and speed of the wave do not change during reflection.
$$\theta_i = \theta_r$$
Boundary Normal Incident Reflected θᵢ θᵣ

3. Wave Refraction & Snell's Law

  • Occurs when a wave crosses a boundary between two media, causing a change in wave speed.
  • Important: Frequency $f$ stays constant! Only the speed $v$ and wavelength $\lambda$ change.
  • When entering a denser medium ($n_2 > n_1$): the wave slows down ($v \downarrow$), wavelength decreases ($\lambda \downarrow$), and it bends towards the normal.
$$n_1 \sin \theta_1 = n_2 \sin \theta_2$$
$$\dfrac{n_1}{n_2} = \dfrac{\sin \theta_2}{\sin \theta_1} = \dfrac{v_2}{v_1} = \dfrac{\lambda_2}{\lambda_1}$$
Medium 1 (n₁) Medium 2 (n₂ > n₁) λ₁ λ₂ Normal θ₁ θ₂

4. Total Internal Reflection (TIR)

  • Occurs only when light travels from a higher refractive index to a lower refractive index.
  • The critical angle $\theta_c$ is defined by $\sin\theta_c = \dfrac{n_1}{n_2} \quad (n_2 > n_1)$.
  • At $\theta_i > \theta_c$, there is no refracted ray; all light is reflected inside the denser medium.
Medium 1 (n₁) Medium 2 (n₂ > n₁) critical ray reflected θᵢ

Example 1: Air-Diamond Boundary

Problem: A light ray is incident on an air–diamond boundary. The refractive index of diamond is greater than $1$. Which diagram shows the correct path of the light ray if Total Internal Reflection occurs?

Air Diamond A

Total internal reflection

Air Diamond B

Without changing direction

Air Diamond C

Diamond to air (toward normal)

Air Diamond D

Air to diamond (away from normal)

Solution:

  • Total internal reflection only happens when light is inside the denser medium (diamond) trying to escape into the less dense medium (air).
  • The incident angle must be greater than the critical angle ($\theta_i > \theta_c$).
  • Because it cannot refract out, all light is perfectly reflected back into the diamond, obeying the law of reflection ($\theta_i = \theta_r$).
  • Therefore, Diagram A shows the correct path of the light ray.

Example 2: Water Waves (Deep to Shallow)

Problem: A water wave entering a harbour passes suddenly from deep to shallow water. In deep water, the wave has frequency $f_1$ and speed $v_1$. In shallow water, the wave has frequency $f_2$ and speed $v_2$.

Deep Water Shallow Water Boundary v₁ v₂

Which of the following compares the frequencies and speeds of the wave between deep water and shallow water?

Solution:

  • Frequency ($f$): The frequency of a wave is determined exclusively by its source. When crossing a boundary into a new medium, the frequency never changes. Therefore, $f_1 = f_2$.
  • Speed ($v$): Water waves naturally travel slower in shallow water than in deep water due to increased interaction with the seabed. Therefore, $v_1 > v_2$.
  • Note: Because $v = f\lambda$ and $f$ is constant, the decrease in speed directly causes the wavelength ($\lambda$) to decrease, which is visually shown by the narrower wavefronts in the shallow region.

Example 3: Light Refraction Diagram

Problem: A light ray passes from air to water as shown.

Air Water Normal Incident Ray

What are the changes in the wavelength of the light wave and the change in the angle of the ray that it makes with the normal to the surface?

Solution:

  • Wavelength: Water is optically denser than air ($n_{\text{water}} > n_{\text{air}}$). When light enters a denser medium, it slows down. Since $v = f\lambda$ and frequency ($f$) remains constant, the wavelength decreases.
  • Angle: Because the wave slows down in the denser medium, the ray bends towards the normal line. Therefore, the angle with the normal decreases ($\theta_2 < \theta_1$).

Example 4: Calculating Relative Refractive Index

Problem: The refractive index for light travelling from medium $X$ to medium $Y$ is $4/3$. The refractive index for light travelling from medium $Y$ to medium $Z$ is $3/5$. What is the refractive index for light travelling from medium $X$ to medium $Z$?

Solution:

  • The relative refractive index for light travelling from medium A to medium B is defined as the ratio of their absolute refractive indices: $_A n_B = \dfrac{n_B}{n_A}$.
  • We are given that
    1) From $X$ to $Y$: $\dfrac{n_Y}{n_X} = \dfrac{4}{3}$
    2) From $Y$ to $Z$: $\dfrac{n_Z}{n_Y} = \dfrac{3}{5}$
  • The refractive index from $X$ to $Z$ is $\dfrac{n_Z}{n_X}$.
  • Multiplying the two given fractions together cancels out $n_Y$:
    $$_{X}n_Z = \left(\dfrac{n_Y}{n_X}\right) \times \left(\dfrac{n_Z}{n_Y}\right) = \dfrac{n_Z}{n_X}$$
  • Substitute the numerical values:
    $$_{X}n_Z = \dfrac{4}{3} \times \dfrac{3}{5} = \dfrac{12}{15} = \mathbf{\dfrac{4}{5}}$$