Reflection, Refraction, and Diffraction

Laws of reflection; Snell's law; critical angle; TIR; diffraction through gaps

# Reflection, Refraction, and Diffraction — IB Physics

When waves meet boundaries or pass through openings, they can reflect, refract, or diffract. These behaviours are fundamental to understanding optics, acoustics, and wave phenomena.


1. Reflection

  • Law of reflection: angle of incidence = angle of reflection (θi=θr\theta_i = \theta_r)
  • Angles measured from the normal
  • Applies to all waves (light, sound, water)

2. Refraction

When a wave passes from one medium to another, its speed changes, causing the wave to bend.

Snell's Law

n1sinθ1=n2sinθ2\boxed{n_1 \sin\theta_1 = n_2 \sin\theta_2}

n=c/vn = c/v (refractive index)

  • Into denser medium: speed decreases, bends towards normal
  • Into less dense medium: speed increases, bends away from normal
  • Frequency remains constant; wavelength changes

3. Total Internal Reflection (TIR)

Conditions:

  1. Light travels from denser to less dense medium
  2. Angle of incidence > critical angle θc\theta_c

sinθc=n2n1\sin\theta_c = \frac{n_2}{n_1}

Applications: optical fibres, prisms, bicycle reflectors.

4. Diffraction

Waves spread out when passing through a gap or around an obstacle.

  • Maximum diffraction when gap width ≈ wavelength
  • Wider gap → less diffraction
  • Longer wavelength → more diffraction

Diffraction explains why we can hear around corners (sound wavelength ~ door width) but light doesn't bend around corners noticeably (very short wavelength).

Worked Example: Snell's Law

Problem

Light passes from air (n=1.00n = 1.00) into glass (n=1.52n = 1.52) at 45°.

sinθ2=(n1/n2)sinθ1=(1/1.52)sin45°=0.465\sin\theta_2 = (n_1/n_2)\sin\theta_1 = (1/1.52)\sin45° = 0.465 θ2=27.7°\theta_2 = 27.7°

Solution

Worked Example: Critical Angle

Problem

Find the critical angle for glass (n=1.50n = 1.50) to air.

sinθc=1/1.50=0.667\sin\theta_c = 1/1.50 = 0.667θc=41.8°\theta_c = 41.8°

Solution

Worked Example: Diffraction

Problem

A doorway is 0.8 m wide. Sound at 425 Hz (v=340v = 340 m/s): λ=0.8\lambda = 0.8 m. Gap ≈ λ\lambda → strong diffraction (sound spreads well). Light at 500 nm: λ\lambda \ll gap → no noticeable diffraction.

Solution

6. Practice Questions

    1. State Snell's law and define refractive index. (2 marks)
    1. Light in water (n=1.33n = 1.33) hits a water-air boundary at 50°. Will TIR occur? (3 marks)
    1. Explain why sound diffracts easily around buildings but light does not. (3 marks)

    Answers

    1. n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2. Refractive index n=c/vn = c/v (ratio of speed of light in vacuum to speed in medium).
    1. θc=sin1(1/1.33)=48.8°\theta_c = \sin^{-1}(1/1.33) = 48.8°. Since 50° > 48.8°: yes, TIR occurs.

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

Summary

  • Reflection: θi=θr\theta_i = \theta_r
  • Refraction: n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2; frequency constant, speed and λ\lambda change
  • TIR: denser → less dense, θ>θc\theta > \theta_c; sinθc=n2/n1\sin\theta_c = n_2/n_1
  • Diffraction: significant when gap ≈ wavelength

Ready to Ace Your IB physics?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store