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Vibrations and waves · Wave phenomena

Phenomena and applications

All waves, whatever their nature, share the same repertoire of behaviours when they meet an obstacle, change medium or overlap with another wave. An echo, a rainbow, wifi passing through a wall and polarised sunglasses are, deep down, the same physics applied to different situations. In this section these phenomena are formalised with the rigour required by the EvAU.

Before you start, make sure you know: Trigonometric ratios of an angle (1st Bachillerato)
Before you start, make sure you know: Light (4th ESO)

Reflection and refraction

When you put a spoon in a glass of water it looks broken right at the surface, and the bottom of a swimming pool seems much closer than it really is: why do light rays bend when they cross the boundary between two transparent substances?

Diagram of an incident ray that is reflected and refracted at the boundary between medium 1 and medium 2, with the angles of incidence, reflection and refraction measured from the normal. normal i r r incident ray reflected ray refracted ray medium 1 (n₁) medium 2 (n₂)
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A ray of light is incident on a plane mirror, making an angle of $35^\circ$ with the normal. The figure shows the incident ray. a) State the two laws of reflection. b)…

A ray of light is incident on a plane mirror, making an angle of $35^\circ$ with the normal. The figure shows the incident ray.

mirror normal i = 35° ?
  1. a) State the two laws of reflection.
  2. b) Determine the angle of reflection and the angle that the reflected ray makes with the surface of the mirror.
Solve
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A ray of light passes from air into water, and it is observed that it bends towards the normal. a) What wave phenomenon has the ray experienced? Define it. b) State…

A ray of light passes from air into water, and it is observed that it bends towards the normal.

  1. a) What wave phenomenon has the ray experienced? Define it.
  2. b) State which wave quantity does not change when passing from one medium to another and which ones do change (propagation speed and wavelength). Justify why the frequency is conserved when changing medium.
Solve
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A ray of light strikes the surface of water ($n_2 = 1,33$) from the air ($n_1 = 1$) with an angle of incidence of $40^\circ$ (data: $\sen 40^\circ \approx 0,64$). The…

A ray of light strikes the surface of water ($n_2 = 1,33$) from the air ($n_1 = 1$) with an angle of incidence of $40^\circ$ (data: $\sen 40^\circ \approx 0,64$). The figure shows the change in direction of the ray.

Ray passing from air to water: the angle of incidence is 40° and the refracted ray bends toward the normal with an unknown angle r air n₁ = 1 water n₂ = 1,33 normal i = 40° r = ?
  1. a) Write Snell's law of refraction and substitute the data from the statement into it, leaving $\sen r$ as the unknown.
  2. b) Calculate the angle of refraction.
Solve

Total internal reflection and the critical angle

Animation of a ray travelling through glass towards air with an angle of incidence that gradually increases: the refracted ray bends away from the normal until it skims along the surface at 90° at the critical angle and then disappears, while the reflected ray becomes stronger until total internal reflection. air (n₂) glass (n₁) i < L r = 90° (limiting case) i > L: total internal reflection
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A glass optical fibre ($n_1 = 1,48$) is surrounded by air ($n_2 = 1$). The figure shows a ray incident on the wall of the fibre at the critical angle. a) Write down…

A glass optical fibre ($n_1 = 1,48$) is surrounded by air ($n_2 = 1$). The figure shows a ray incident on the wall of the fibre at the critical angle.

glass n₁ = 1,48 air n₂ = 1 i = ? r = 90°
  1. a) Write down Snell's law for a ray incident on the glass-air interface at the critical angle, for which the angle of refraction is $90^\circ$.
  2. b) Calculate the critical angle of the fibre.
Solve
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A diver points their torch towards the surface of the water ($n_{\text{agua}} = 1,33$; $n_{\text{aire}} = 1$) at an angle of $50^\circ$ to the vertical. The diagram…

A diver points their torch towards the surface of the water ($n_{\text{agua}} = 1,33$; $n_{\text{aire}} = 1$) at an angle of $50^\circ$ to the vertical. The diagram represents the situation.

air n = 1 water n = 1,33 torch i = 50° ?
  1. a) Calculate the critical angle for the transition from water to air.
  2. b) Explain whether the beam from the torch will emerge into the air or undergo total internal reflection, by comparing the angle of incidence with the critical angle.
Solve
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An optical fibre has a core of index $n_1 = 1,50$ and a cladding of index $n_2 = 1,47$. The diagram shows a section of fibre with a guided ray. a) Calculate the critical…

An optical fibre has a core of index $n_1 = 1,50$ and a cladding of index $n_2 = 1,47$. The diagram shows a section of fibre with a guided ray.

n₂ = 1,47 cladding n₁ = 1,50 core critical angle = ?
  1. a) Calculate the critical angle at the core-cladding interface.
  2. b) Explain why light propagates through the core without escaping when incident on its walls at angles greater than this critical angle.
  3. c) If the core were in direct contact with air ($n = 1$), calculate the new critical angle and comment on the practical advantage provided by the cladding.
Solve

Interference of waves

In a room with two identical loudspeakers emitting the same pure tone, there are areas where the sound is heard very loudly and others a few steps away where you can hardly hear anything: how can two sound sources cancel each other out when they meet at a point?

Two coherent sound sources emitting towards a point P, situated at distances d₁ and d₂ from each source; the path difference Δr = d₂ − d₁ determines the type of interference. source 1 source 2 P d₁ d₂ Δr = d₂ − d₁
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Two loudspeakers emit the same note in phase. A microphone placed on the perpendicular bisector of the line segment joining them is at the same distance from both. The…

Two loudspeakers emit the same note in phase. A microphone placed on the perpendicular bisector of the line segment joining them is at the same distance from both. The diagram shows the arrangement of the loudspeakers and the microphone.

loudspeaker 1 loudspeaker 2 perpendicular bisector microphone d₁ d₂
  1. a) With what path difference do the two waves arrive at the microphone?
  2. b) Reason whether the interference at that point is constructive or destructive and what will be heard there.
Solve
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Two in-phase sources emit waves of wavelength $\lambda = 2\ \text{m}$. At a point P, the distances to the sources are $d_1 = 10\ \text{m}$ and $d_2 = 13\ \text{m}$. The…

Two in-phase sources emit waves of wavelength $\lambda = 2\ \text{m}$. At a point P, the distances to the sources are $d_1 = 10\ \text{m}$ and $d_2 = 13\ \text{m}$. The figure shows the two sources and point P.

source 1 source 2 λ = 2 m d₁ = 10 m d₂ = 13 m P
  1. a) Calculate the path difference $\Delta r = d_2 - d_1$.
  2. b) Determine whether $\Delta r$ is an integer number of wavelengths or an odd number of half-wavelengths, and conclude whether the interference at P is constructive or destructive.
Solve
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In an optics laboratory, a double slit is illuminated with a red laser of $\lambda = 633\ \text{nm} = 6,33 \cdot 10^{-7}\ \text{m}$. The slits are separated by…

In an optics laboratory, a double slit is illuminated with a red laser of $\lambda = 633\ \text{nm} = 6,33 \cdot 10^{-7}\ \text{m}$. The slits are separated by $d = 0,2\ \text{mm} = 2 \cdot 10^{-4}\ \text{m}$ and the screen is at $L = 2\ \text{m}$. In the small-angle approximation, the distance between two consecutive bright fringes is $\Delta y = \frac{\lambda \cdot L}{d}$ (given formula). The diagram summarises the set-up, not to scale.

laser λ = 633 nm slits d = 0,2 mm screen L = 2 m Δy = ?
  1. a) Calculate $\Delta y$ and express it in millimetres.
  2. b) Explain the origin of the fringes as interference of the waves coming from the two slits: what condition does the path difference satisfy at a bright fringe and at a dark fringe?
  3. c) Predict how the pattern will change if the red laser is replaced by a blue one, of shorter wavelength.
Solve

Diffraction

Diffraction: a plane wave strikes a narrow slit in a barrier and, on passing through it, spreads out into circular wavefronts that propagate beyond the straight-line region. slit incident wave the wave spreads out beyond the slit no diffraction
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From the hallway, the living room television can be heard even though the screen cannot be seen. The floor plan of the house outlines the situation. a) What wave…

From the hallway, the living room television can be heard even though the screen cannot be seen. The floor plan of the house outlines the situation.

living room hallway doorway television person
  1. a) What wave phenomenon allows sound to "bend round the corner"?
  2. b) Explain, by comparing the width of the doorway with the typical wavelengths of sound (of the order of $1\ \text{m}$) and light (of the order of $5 \cdot 10^{-7}\ \text{m}$), why sound diffracts appreciably at the doorway while light does not.
Solve
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When tilting a CD under white light, rainbow-like colours appear. a) Explain the role of the microscopic grooves on the disc, which act as a reflection diffraction…

When tilting a CD under white light, rainbow-like colours appear.

  1. a) Explain the role of the microscopic grooves on the disc, which act as a reflection diffraction grating.
  2. b) Explain why each colour is observed in a different direction, relating it to the constructive interference of each wavelength and to $\lambda_{\text{rojo}} > \lambda_{\text{azul}}$.
Solve
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A home router emits its Wi-Fi signal in the $2,4\ \text{GHz}$ band (given: $c = 3 \cdot 10^{8}\ \text{m/s}$). a) Calculate the wavelength of the signal and express it in…

A home router emits its Wi-Fi signal in the $2,4\ \text{GHz}$ band (given: $c = 3 \cdot 10^{8}\ \text{m/s}$).

  1. a) Calculate the wavelength of the signal and express it in centimetres.
  2. b) Compare it with the width of a doorway ($0,8\ \text{m}$) and explain, naming the wave phenomenon responsible, why there is coverage in the adjacent room even though there is no direct line of sight to the router.
Solve

Polarisation

If you place two polarised sunglass lenses on top of each other and rotate one of them by $90^\circ$, light stops passing through completely and the pair becomes totally opaque: what geometric orientation of the light wave allows it to be blocked with a simple rotation?

Natural light, which vibrates in all directions perpendicular to the direction of propagation, passes through a polarising filter and emerges vibrating only vertically. natural light: vibrates in all directions filter polariser vibrates only vertically

Back to the start: An echo is the reflection of sound bouncing off an obstacle, a rainbow comes from the refraction and internal reflection of sunlight in raindrops, wifi reaches rooms with no line of sight to the router partly thanks to the diffraction of its waves through doorways and openings (as well as passing through the walls), and sunglasses block glare because light reflected from the ground is partially polarised.

Summary

See all the exercises in this section