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 n₁ sin θ₁ = n₂ sin θ₂

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

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 4πr². This sphere area is not given, so learn it. With no absorption,

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/r², because the medium absorbs some of the energy.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) A = 4π×8.0² = 804 m², so I = 2.5 ÷804 = 3.1 × 10⁻³ W m⁻².

(b) Twice the distance, a quarter of the intensity: 3.11 ×10⁻³ ÷4 = 7.8 × 10⁻⁴ W m⁻².

(c) r = √(2.5 ÷(4π×1.0 ×10⁻⁵)) = 141 m ≈ 140 m.

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 = fλ.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) 0.040 ×850 = 34 W m⁻².

(b) 96% of what arrives is transmitted at each surface: fraction transmitted = 0.96⁴ = 0.849, so I = 0.849 ×850 = 722 W m⁻². Power = intensity × area: P = 722 ×1.8 = 1.3 kW (1300 W).

Subtracting 4 ×4.0% = 16% would be wrong: each surface reflects 4.0% of a slightly smaller intensity.

Here c = 3.00 ×10⁸ m s⁻¹, v is the speed of light in the material, and θ₁, θ₂ 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 (θ= 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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Snell's law with n₁ = 1.00: sin θ₂ = sin 52° ÷1.52 = 0.788 ÷1.52 = 0.518, so θ₂ = 31.2°.

(b) v = c/n = 3.00 ×10⁸ ÷1.52 = 1.97 × 10⁸ m s⁻¹.

(c) The frequency is unchanged, so the wavelength falls by the same factor as the speed: λ= 633 ÷1.52 = 416 nm.

Check: f = c/λ= 4.74 ×10¹⁴ Hz in air, and (1.97 ×10⁸) ÷(416 ×10⁻⁹) gives the same frequency in glass.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) 1.33 sin 40° = 1.52 sin θ_g, so sin θ_g = 0.855 ÷1.52 = 0.562 and θ_g = 34.2°.

(b) With parallel faces, the angle of incidence at the glass–air face is 34.2°: 1.52 ×0.562 = 1.00 ×sin θₐ, so sin θₐ = 0.855 and θₐ = 58.7°.

The product nsin θ is the same in every layer, so a parallel-sided wall has no effect on the final direction: 1.33 sin 40° = sin θₐ directly.

sin C = 1/n is in the formulae list supplied in the exam. Between two materials, put θ₂ = 90° into Snell's law to get sin C = n₂/n₁ with n₁ > n₂; 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_core and n_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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) sin C = 1.48 ÷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 θₐᵢᵣ = 1.55 ×sin 17.3° = 0.461, so θₐᵢᵣ = 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.

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.

WORKED EXAMPLE · QUESTION

A student measures these angles for a glass block, to the nearest degree.

(a) Use a graph to find the refractive index of the glass. (b) Compare the percentage uncertainties in r = 13° and r = 39°.

SOLUTION

(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 ÷0.50 = n = 1.50.

(b) Uncertainty = half the 1° resolution = ±0.5°. 13°: 0.5 ÷13 ×100 = 3.8%; 39°: 0.5 ÷39 ×100 = 1.3%.

The large-angle reading is about three times more precise.

A single reading gives only one estimate, for example sin 70° ÷sin 39° = 1.49; the gradient uses all six.

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.

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

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 τ is vτ long. Two reflectors closer together than vτ/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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) The echo time is for there and back, so halve it: d_A = 1540 ×26 ×10⁻⁶ ÷2 = 2.0 cm and d_B = 1540 ×91 ×10⁻⁶ ÷2 = 7.0 cm. Separation = 7.0 - 2.0 = 5.0 cm.

(b) λ= v/f = 1540 ÷3.5 ×10⁶ = 0.44 mm.

(c) Pulse length in tissue = 1540 ×1.4 ×10⁻⁶ = 2.16 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.

1 Progressive waves and 2.4 Diffraction

v = fλ 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.

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/4πd², is the same inverse-square law, used to find stellar distances and luminosities.

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.

2. Light travels at 2.26 × 10⁸ m s⁻¹ in a liquid. Calculate the refractive index of the liquid.

3. Calculate the critical angle for light going from glass of refractive index 1.60 into air.

4. Explain why sound waves cannot be polarised but light waves can.

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.

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?

1. At 4.0 m: 8.0 ×(2.0/4.0)² = 2.0 W m⁻². At 6.0 m: 8.0 ×(2.0/6.0)² = 0.89 W m⁻².

2. n = c/v = 3.00 ×10⁸ ÷2.26 ×10⁸ = 1.33.

3. sin C = 1/1.60 = 0.625, so C = 38.7°.

4. 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. d = vt/2 = 1500 ×0.20 ÷2 = 150 m.

6. sin C = 1.33 ÷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.