2.3 Intensity, refraction and polarisation
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AS content for Unit 2 (Waves and Electricity). Wave basics are on 2.1, superposition on 2.2 and diffraction on 2.4. Malus's law is not needed.
What this page covers
7 points
- Intensity as power per unit area, and the inverse-square law for a point source
- Partial reflection and transmission of a wave at a boundary between two media
- Refractive index, \(n = c/v\), and Snell's law \(n_1 \sin\theta_1 = n_2 \sin\theta_2\)
- Critical angle and predicting total internal reflection, including optical fibres
- Measuring the refractive index of a glass block from a graph of sin i against sin r
- Plane polarisation, polarising filters and their uses
- The pulse-echo technique and what limits the detail it can show
In one minute
6 points
- Intensity is power per unit area, \(I = P/A\) in W m⁻², and from a point source \(I = P/4\pi r^2\), so doubling the distance gives a quarter of the intensity.
- At a boundary a wave is partly reflected and partly transmitted; its frequency stays the same while its speed and wavelength change.
- Refractive index \(n = c/v\). Across a boundary, \(n_1 \sin\theta_1 = n_2 \sin\theta_2\), with angles measured from the normal.
- Total internal reflection needs light travelling towards a lower refractive index and an angle of incidence greater than the critical angle, \(\sin C = 1/n\).
- Plane-polarised waves oscillate in one plane only. Only transverse waves can be polarised.
- Pulse-echo: distance \(= vt/2\). Detail is limited by the wavelength and by the pulse duration.
Contents
13 sections
- What is intensity, and how does it fall with distance?
- What happens when a wave reaches a boundary between two media?
- How do you use refractive index and Snell's law?
- When does total internal reflection happen?
- How do you measure the refractive index of a glass block?
- What is plane polarisation, and what is it used for?
- How does the pulse-echo technique locate objects?
- Exam tips for 2.3
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is intensity, and how does it fall with distance?
A lamp seems dimmer and a siren quieter the further away you are, even though the source puts out the same power. What changes is how concentrated that power is when it reaches you.
\(I = P/A\) is in the formulae list supplied in the exam. A point source radiating equally in all directions spreads its power over the surface of a sphere, of area \(4\pi r^2\). This sphere area is not given, so learn it. With no absorption,
\[I = \frac{P}{4\pi r^2} \qquad\text{so}\qquad I \propto \frac{1}{r^2}\]This is the inverse-square law. Doubling the distance spreads the same power over four times the area, because both dimensions of any patch double; the intensity falls to a quarter. Trebling the distance gives a ninth.
- A beam that does not spread (an ideal laser beam) keeps the same area, so its intensity does not fall with distance.
- Real intensities often fall faster than \(1/r^2\), because the medium absorbs some of the energy.
A siren on a factory roof emits sound with a power of 2.5 W, equally in all directions. Ignore absorption and reflections. (a) Calculate the intensity 8.0 m from the siren. (b) State the intensity at 16 m. (c) Calculate the distance at which the intensity has fallen to 1.0 × 10⁻⁵ W m⁻².
(a) \(A = 4\pi \times 8.0^2 = 804\ \text{m}^2\), so \(I = 2.5 \div 804 =\) 3.1 × 10⁻³ W m⁻².
(b) Twice the distance, a quarter of the intensity: \(3.11 \times 10^{-3} \div 4 =\) 7.8 × 10⁻⁴ W m⁻².
(c) \(r = \sqrt{2.5 \div (4\pi \times 1.0 \times 10^{-5})} = 141\ \text{m} \approx\) 140 m.
Using \(\pi r^2\) (the area of a circle) instead of \(4\pi r^2\) (the surface area of a sphere) for a source that radiates in all directions. The answer is then four times too big.
What happens when a wave reaches a boundary between two media?
At a boundary, part of the wave is reflected and part is transmitted into the new medium; some energy may also be absorbed. Ignoring absorption, energy is conserved: incident intensity = reflected intensity + transmitted intensity.
- The more different the two media (for light, the bigger the difference in refractive index; for sound and ultrasound, the bigger the difference in density and wave speed), the larger the fraction reflected.
- The frequency does not change: it is set by the source, and every wavefront that arrives must leave. The speed changes, so the wavelength changes too, since \(v = f\lambda\).
- If the wave meets the boundary at an angle and its speed changes, its direction changes: this is refraction.
- Almost all ultrasound is reflected at an air–skin boundary, so a gel is spread on the skin to remove the air layer and let the ultrasound in.
Sunlight of intensity 850 W m⁻² falls at right angles on a double-glazed window of area 1.8 m². The window has two panes, so the light crosses four glass–air surfaces. At each surface 4.0% of the intensity arriving there is reflected. Ignore absorption in the glass. (a) Calculate the intensity reflected at the first surface. (b) Calculate the power of the sunlight that reaches the room.
(a) \(0.040 \times 850 =\) 34 W m⁻².
(b) 96% of what arrives is transmitted at each surface: fraction transmitted \(= 0.96^4 = 0.849\), so \(I = 0.849 \times 850 = 722\ \text{W m}^{-2}\). Power = intensity × area: \(P = 722 \times 1.8 =\) 1.3 kW (1300 W).
Subtracting \(4 \times 4.0\% = 16\%\) would be wrong: each surface reflects 4.0% of a slightly smaller intensity.
How do you use refractive index and Snell's law?
Here \(c = 3.00 \times 10^{8}\ \text{m s}^{-1}\), \(v\) is the speed of light in the material, and \(\theta_1, \theta_2\) are the angles to the normal in media 1 and 2. Both equations are in the formulae list supplied in the exam. Take \(n = 1.00\) for air, and always measure angles from the normal, never from the surface.
- \(n\) has no unit and is at least 1. A larger \(n\) means slower light (an "optically denser" material).
- Entering a medium of higher \(n\), light slows and bends towards the normal; entering a lower \(n\), it bends away from the normal.
- A ray along the normal (\(\theta = 0\)) does not bend, although it still slows down.
- Through a parallel-sided block, the ray leaves parallel to its original direction but shifted sideways.
A beam of red laser light of wavelength 633 nm in air strikes a glass surface at 52° to the normal. The refractive index of the glass is 1.52. Calculate (a) the angle of refraction, (b) the speed of the light in the glass and (c) its wavelength in the glass.
(a) Snell's law with \(n_1 = 1.00\): \(\sin\theta_2 = \sin 52° \div 1.52 = 0.788 \div 1.52 = 0.518\), so \(\theta_2 =\) 31.2°.
(b) \(v = c/n = 3.00 \times 10^{8} \div 1.52 =\) 1.97 × 10⁸ m s⁻¹.
(c) The frequency is unchanged, so the wavelength falls by the same factor as the speed: \(\lambda = 633 \div 1.52 =\) 416 nm.
Check: \(f = c/\lambda = 4.74 \times 10^{14}\) Hz in air, and \((1.97 \times 10^{8}) \div (416 \times 10^{-9})\) gives the same frequency in glass.
Light travelling in the water of an aquarium (\(n = 1.33\)) meets the glass wall (\(n = 1.52\)) at 40° to the normal. The wall has parallel faces and the light passes through it into the air. Calculate (a) the angle of refraction in the glass and (b) the angle at which the light leaves into the air.
(a) \(1.33 \sin 40° = 1.52 \sin\theta_g\), so \(\sin\theta_g = 0.855 \div 1.52 = 0.562\) and \(\theta_g =\) 34.2°.
(b) With parallel faces, the angle of incidence at the glass–air face is 34.2°: \(1.52 \times 0.562 = 1.00 \times \sin\theta_a\), so \(\sin\theta_a = 0.855\) and \(\theta_a =\) 58.7°.
The product \(n\sin\theta\) is the same in every layer, so a parallel-sided wall has no effect on the final direction: \(1.33 \sin 40° = \sin\theta_a\) directly.
When does total internal reflection happen?
\(\sin C = 1/n\) is in the formulae list supplied in the exam. Between two materials, put \(\theta_2 = 90°\) into Snell's law to get \(\sin C = n_2/n_1\) with \(n_1 > n_2\); derive this, as it is not given. Total internal reflection (TIR) happens only when both conditions hold:
- the light is travelling from a medium of higher refractive index towards one of lower refractive index, and
- the angle of incidence is greater than the critical angle.
Below C there is refraction out plus a weak partial reflection. At C the refracted ray runs along the surface. Above C all the light is reflected. To predict TIR, calculate C for that pair of materials and compare it with the actual angle of incidence.
Optical fibres
A thin glass core is surrounded by cladding of slightly lower refractive index. Light hitting the core–cladding boundary above the critical angle is totally internally reflected and guided along the fibre, even round gentle bends. The cladding keeps the core surface clean and unscratched, stops light leaking into a touching fibre, and keeps the boundary condition the same all along the fibre. Because \(n_\text{core}\) and \(n_\text{cladding}\) are close, the critical angle is large, so only rays nearly parallel to the axis are guided. Uses include data communication and endoscopes that carry images out of the body.
An optical fibre has a core of refractive index 1.55 and cladding of refractive index 1.48. The end face of the fibre is flat and perpendicular to its axis, with air outside. (a) Calculate the critical angle at the core–cladding boundary. (b) A ray meets this boundary at 76° to the normal. State and explain what happens. (c) Calculate the largest angle a ray in the core can make with the fibre axis and still be guided. (d) Hence find the largest angle to the axis at which light in the air can enter the end face and be guided.
(a) \(\sin C = 1.48 \div 1.55 = 0.955\), so C = 72.7°.
(b) 76° > 72.7°, and the light is going from higher to lower \(n\), so total internal reflection.
(c) The boundary is parallel to the axis, so the angle to the axis is 90° minus the angle of incidence: \(90° - 72.7° =\) 17.3°.
(d) Snell's law at the end face, air → core (the normal is the axis): \(\sin\theta_\text{air} = 1.55 \times \sin 17.3° = 0.461\), so \(\theta_\text{air} =\) 27.4°.
A ray meeting the boundary at 70° would be partly transmitted into the cladding at every reflection and the signal would fade. Using \(1/1.55\) would give 40.2°, the critical angle for core to air, which is not the boundary the light meets.
How do you measure the refractive index of a glass block?
- Set-up: place a rectangular glass block on paper and draw round it. Use a ray box with a single narrow slit to send a thin ray at the long face.
- Recording rays: mark the incident and emergent rays with small pencil crosses, well apart. Remove the block and join the entry and exit points: this line is the refracted ray inside the glass.
- Angles: draw the normal at the entry point and measure \(i\) and \(r\) from the normal with a protractor. Repeat for at least six angles of incidence, for example 20° to 70°.
- Analysis: with air outside, \(\sin i = n \sin r\). A graph of \(\sin i\) (y-axis) against \(\sin r\) (x-axis) is a straight line through the origin with gradient \(n\).
- Precision: a protractor reading to 1° has an uncertainty of ±0.5°, a smaller percentage of a larger angle, so large angles are more reliable. A thin ray and a sharp pencil help.
- Safety: the ray box lamp gets hot; handle it by its case and let it cool.
A student measures these angles for a glass block, to the nearest degree.
| i / ° | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|
| r / ° | 13 | 20 | 25 | 31 | 35 | 39 |
(a) Use a graph to find the refractive index of the glass. (b) Compare the percentage uncertainties in r = 13° and r = 39°.
(a) sin r: 0.225, 0.342, 0.423, 0.515, 0.574, 0.629. sin i: 0.342, 0.500, 0.643, 0.766, 0.866, 0.940. Plot sin i against sin r, draw the best-fit line and find its gradient from a large triangle: gradient \(= 0.749 \div 0.50 =\) n = 1.50.
(b) Uncertainty = half the 1° resolution = ±0.5°. 13°: \(0.5 \div 13 \times 100 = 3.8\%\); 39°: \(0.5 \div 39 \times 100 = 1.3\%\). The large-angle reading is about three times more precise.
A single reading gives only one estimate, for example \(\sin 70° \div \sin 39° = 1.49\); the gradient uses all six.
What is plane polarisation, and what is it used for?
- In unpolarised light (from the Sun or a filament lamp) the oscillations occur in all directions perpendicular to the direction of travel.
- A polarising filter transmits only the oscillations along its transmission axis, so unpolarised light leaving it is plane-polarised.
- Two filters with axes at 90° ("crossed") let no light through. Rotating the second filter through 360° gives two bright and two dark positions.
- Only transverse waves can be polarised. In a longitudinal wave such as sound the oscillations are along the direction of travel, so there is no plane to select. That light can be polarised is evidence that light is transverse.
- Microwaves from a transmitter horn are already plane-polarised: turning the receiver through 90° about the beam makes the signal fall to a minimum.
Uses of polarisation
- Sunglasses: light reflected from water, wet roads and glass is partly plane-polarised, mostly in a horizontal plane. Polarising sunglasses with vertical transmission axes absorb much of this glare while passing other light.
- Stress analysis: a transparent plastic model of a part is placed between crossed polarisers. Stressed plastic rotates the plane of polarisation by different amounts for different colours, so coloured fringes appear; closely spaced fringes show where the stress is concentrated.
- Displays and aerials: liquid-crystal displays use polarising filters, and a TV or radio aerial must be lined up with the plane of polarisation of the transmitted waves for the best signal.
Defining polarisation as "the wave vibrates in one direction". Say oscillations in one plane only, and that the plane includes the direction of travel.
How does the pulse-echo technique locate objects?
A short pulse is sent out; part of it reflects from each boundary or object it meets and returns as an echo to a detector next to the source. The time \(t\) between sending the pulse and receiving the echo is for the journey there and back, so the distance to the reflector is
\[d = \frac{vt}{2}\]This is not in the formulae list: derive it from the fact that the pulse goes there and back. Ultrasound scans use frequencies of a few MHz in the body; sonar uses sound or ultrasound in water to find the sea bed, fish or submarines; radar uses radio waves or microwaves, travelling at the speed of light, to find aircraft and ships. An echo trace (an A-scan) plots the detected signal against time: each spike is a boundary, and its delay gives the depth. Echoes get weaker with depth, because each boundary reflects only part of the pulse and the medium absorbs energy.
What limits the detail?
- Wavelength: a feature much smaller than the wavelength does not give a clear reflection, because the wave diffracts round it. A shorter wavelength, from a higher frequency, shows finer detail.
- Pulse duration: a pulse of duration \(\tau\) is \(v\tau\) long. Two reflectors closer together than \(v\tau/2\) along the beam send back echoes that overlap and look like one. Shorter pulses separate closer boundaries.
- Higher frequency also lets each pulse contain the same number of cycles in a shorter time, but high frequencies are absorbed more, so they cannot reach as deep.
The trace above comes from a 3.5 MHz ultrasound probe on the skin. The speed of ultrasound in soft tissue is 1540 m s⁻¹, and each pulse lasts 1.4 μs. (a) Calculate the depths of boundaries A and B and the distance between them. (b) Calculate the wavelength of the ultrasound in tissue. (c) Calculate the smallest separation of two boundaries, along the beam, whose echoes do not overlap.
(a) The echo time is for there and back, so halve it: \(d_A = 1540 \times 26 \times 10^{-6} \div 2 = 2.0\ \text{cm}\) and \(d_B = 1540 \times 91 \times 10^{-6} \div 2 = 7.0\ \text{cm}\). Separation \(= 7.0 - 2.0 =\) 5.0 cm.
(b) \(\lambda = v/f = 1540 \div 3.5 \times 10^{6} =\) 0.44 mm.
(c) Pulse length in tissue \(= 1540 \times 1.4 \times 10^{-6} = 2.16\ \text{mm}\); halve it for the round trip: 1.1 mm.
The echo from B is weaker: some of the pulse was reflected at A, and the tissue absorbs energy on the longer path.
Exam tips for 2.3
Every angle in Snell's law and in the critical-angle condition is measured from the normal. If a question gives the angle to the surface, subtract it from 90° first. Write \(n_1 \sin\theta_1 = n_2 \sin\theta_2\) with symbols, label media 1 and 2, and sense-check: going into a higher \(n\), the angle must get smaller.
State both: travelling towards a lower refractive index, and an angle of incidence greater than the critical angle. Calculate C and quote it in the comparison.
At a boundary the frequency is fixed by the source. Find the new speed from \(n = c/v\), then the new wavelength from \(\lambda = v/f\), or divide the vacuum wavelength by \(n\).
For a point source, compare two distances with \(I_2/I_1 = (r_1/r_2)^2\); there is no need to know the power. In every echo calculation write "time there and back, so distance \(= vt/2\)" as a line of working.
When asked what limits the detail in a pulse-echo image, give both the wavelength (features smaller than it are not resolved) and the pulse duration (boundaries closer than \(v\tau/2\) give overlapping echoes). In the glass-block practical, say which quantity goes on which axis: with sin i on the y-axis the gradient is \(n\).
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Doubling the distance from a source halves the intensity. | The power spreads over a sphere whose area grows as \(r^2\). | Doubling the distance quarters the intensity (point source, no absorption). |
| The frequency of light changes when it enters glass. | Frequency is set by the source; every wavefront that arrives must leave. | Speed and wavelength change; frequency stays the same. |
| Light bends towards the normal whenever it crosses a boundary. | The direction depends on whether it speeds up or slows down. | Towards the normal into higher \(n\); away from the normal into lower \(n\). |
| Total internal reflection can happen when light enters glass from air. | Going into a higher \(n\), \(\sin\theta_2\) is always less than 1, so a refracted ray always exists. | TIR needs travel towards a lower refractive index and an angle above C. |
| Below the critical angle all the light is transmitted. | Every boundary reflects part of the wave. | Below C there is refraction plus a weak partial reflection. |
| Sound can be polarised with the right filter. | Its oscillations are along the direction of travel, so there is no plane to select. | Only transverse waves can be polarised. |
How this topic connects
2.1 Progressive waves and 2.4 Diffraction
\(v = f\lambda\) turns a change of speed at a boundary into a change of wavelength, and the transverse–longitudinal distinction is exactly what polarisation tests. Diffraction (2.4) is why the wavelength limits the detail of an ultrasound or sonar image.
2.5 Photons and the photoelectric effect
Intensity reappears in the photoelectric effect: a more intense beam of the same frequency delivers more photons per second, not more energetic ones.
Unit 5 astrophysics
The flux from a star, \(F = L/4\pi d^2\), is the same inverse-square law, used to find stellar distances and luminosities.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(I = P/A\) | intensity from power and area; I in W m⁻², P in W, A in m² | Yes |
| \(n = c/v\) | refractive index from speeds; no unit | Yes |
| \(n_1 \sin\theta_1 = n_2 \sin\theta_2\) | refraction at a boundary; angles from the normal | Yes |
| \(\sin C = 1/n\) | critical angle, material to air | Yes |
| \(v = f\lambda\) | wavelength change at a boundary (earlier page) | Yes |
| \(A = 4\pi r^2\), so \(I = P/4\pi r^2\) | point source radiating in all directions | No — learn the sphere area |
| \(\sin C = n_2/n_1\) | critical angle between two materials | No — derive it: put θ₂ = 90° in Snell's law |
| \(d = vt/2\) | pulse-echo distance | No — derive it: the pulse goes there and back |
Check your understanding
1. The intensity 2.0 m from a small lamp is 8.0 W m⁻². Calculate the intensity at 4.0 m and at 6.0 m.
At 4.0 m: \(8.0 \times (2.0/4.0)^2 =\) 2.0 W m⁻². At 6.0 m: \(8.0 \times (2.0/6.0)^2 =\) 0.89 W m⁻².
2. Light travels at 2.26 × 10⁸ m s⁻¹ in a liquid. Calculate the refractive index of the liquid.
\(n = c/v = 3.00 \times 10^{8} \div 2.26 \times 10^{8} =\) 1.33.
3. Calculate the critical angle for light going from glass of refractive index 1.60 into air.
\(\sin C = 1/1.60 = 0.625\), so C = 38.7°.
4. Explain why sound waves cannot be polarised but light waves can.
Sound is longitudinal: its oscillations are along the direction of travel, so there is no plane to select. Light is transverse, with oscillations perpendicular to its direction of travel that can be restricted to one plane.
5. A ship's sonar pulse returns from the sea bed 0.20 s after it is sent. Sound travels at 1500 m s⁻¹ in sea water. Calculate the depth.
\(d = vt/2 = 1500 \times 0.20 \div 2 =\) 150 m.
6. Light in glass (n = 1.52) meets a boundary with water (n = 1.33) at 65° to the normal. Does total internal reflection occur?
\(\sin C = 1.33 \div 1.52 = 0.875\), so \(C = 61.0°\). The light travels towards a lower refractive index and 65° > 61.0°, so yes, total internal reflection occurs.
Specification checklist
Numbers refer to the specification points.
- (44) I can use \(I = P/A\), including for a point source whose power spreads over a sphere of area \(4\pi r^2\).
- (45) I can use \(n = c/v\) and \(n_1 \sin\theta_1 = n_2 \sin\theta_2\), with angles measured from the normal.
- (46) I can calculate a critical angle from \(\sin C = 1/n\), and between two materials.
- (47) I can predict whether total internal reflection will happen, and explain how an optical fibre guides light.
- (48) I can describe how to find the refractive index of a glass block using a ray box and protractor, and a graph of sin i against sin r.
- (49) I can explain plane polarisation, why only transverse waves can be polarised, and uses such as sunglasses and stress analysis.
- (55) I can explain that a wave meeting a boundary is partly reflected and partly transmitted, with unchanged frequency.
- (56) I can use the pulse-echo technique to find distances, and explain how wavelength and pulse duration limit the detail.