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5th Year HL · Physics
Heat Transfer, Irradiance, Diffraction and 2D Collisions · Irradiance and electromagnetic energy sources

Irradiance

A solar panel on a roof in Cork and a solar panel on a roof in Athens are the same size, yet they do not produce the same amount of electricity. Both are hit by sunlight, so what exactly are we comparing? The answer is a quantity called irradiance, and by the end of this page you will be able to explain why the Cork panel gets less.

What irradiance means

Power tells you how much energy a source delivers each second. But a beam of light spread over a big area feels weaker than the same beam squeezed onto a small one. To compare fairly, we share the power over the area it falls on.

Notice that the same power on half the area doubles the irradiance. This is why a magnifying glass can set paper alight: it does not create energy, it concentrates it.

The same radiant power spread over a large area gives a low irradiance, and over a small area gives a high irradiance. Power 100 W on 2 m² I = 50 W m⁻² Power 100 W on 1 m² I = 100 W m⁻²

Once you have the irradiance, you can also work backwards. Power is $P = IA$, and energy is $E = Pt$. In everyday life energy is often given in kilowatt-hours: $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$.

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Irradiance is defined by $I=\dfrac{P}{A}$. a) State the meaning of each symbol and the SI unit of irradiance. b) A solar panel of area $1.6\ \text{m}^2$ receives…

Irradiance is defined by $I=\dfrac{P}{A}$.

  1. a) State the meaning of each symbol and the SI unit of irradiance.
  2. b) A solar panel of area $1.6\ \text{m}^2$ receives radiation of power $1,200\ \text{W}$. Calculate the irradiance.
  3. c) Explain what happens to the irradiance on a surface if the area is halved while the power is unchanged.
Solve
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On a clear day in Cork the irradiance of sunlight on a horizontal surface is $800\ \text{W m}^{-2}$. a) Calculate the power incident on a roof surface of area…

On a clear day in Cork the irradiance of sunlight on a horizontal surface is $800\ \text{W m}^{-2}$.

  1. a) Calculate the power incident on a roof surface of area $25\ \text{m}^2$.
  2. b) Calculate the energy incident on this surface in $2.0$ hours.
  3. c) Express your answer to part b) in $\text{kWh}$.
Solve

Efficiency and solar farms

A solar panel does not turn all the incident power into electricity. Efficiency compares the useful output with the total input:

$$\text{efficiency} = \frac{\text{useful power out}}{\text{power in}} \times 100\%$$

The power in is $P = IA$. So the electrical power is $P_{out} = \text{efficiency} \times IA$.

Why is the average irradiance so much lower than $800\ \text{W m}^{-2}$ on a clear day? Because the average includes nights, cloud and low winter sun.

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A solar farm in County Wexford has panels of total area $4,000\ \text{m}^2$. The average irradiance over a year is $120\ \text{W m}^{-2}$ and the panels are $18\%$…

A solar farm in County Wexford has panels of total area $4,000\ \text{m}^2$. The average irradiance over a year is $120\ \text{W m}^{-2}$ and the panels are $18\%$ efficient.

  1. a) Calculate the average power incident on the panels.
  2. b) Calculate the average electrical power produced.
  3. c) Calculate the electrical energy produced in one year, in $\text{kWh}$.
  4. d) An average Irish household uses about $4,200\ \text{kWh}$ per year. Estimate how many households the farm could supply.
Solve

Distance: the inverse-square law

Stand beside a lamp and it is bright. Walk away and it dims quickly. The lamp's total power has not changed, but it now spreads over a bigger area.

A point source emits equally in all directions. At distance $r$ the energy is spread over a sphere of area $4\pi r^2$, so

$$I = \frac{P}{4\pi r^2}$$ A point source with the same power spreading over spheres of radius r and 2r; the larger sphere has four times the area. Inner circle: radius r, area 4πr² Dashed circle: radius 2r, area 16πr² (four times larger)

Doubling the distance makes the area 4 times bigger, so the irradiance falls to a quarter. This is an inverse square law: $I \propto \dfrac{1}{r^2}$.

This is also how you verify the law with a light meter: measure $I$ at several distances $d$, and show that $Id^2$ is the same each time, or that a graph of $I$ against $1/d^2$ is a straight line through the origin.

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A point source of light, a lamp, emits radiation uniformly in all directions with a total power of $60\ \text{W}$. a) Calculate the irradiance at a distance of…

A point source of light, a lamp, emits radiation uniformly in all directions with a total power of $60\ \text{W}$.

  1. a) Calculate the irradiance at a distance of $1.0\ \text{m}$ from the lamp, using the surface area of a sphere $A=4\pi r^2$.
  2. b) Calculate the irradiance at $3.0\ \text{m}$.
  3. c) State the relationship between irradiance and distance for a point source, and explain it with reference to the spreading of energy.
Solve
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A student measures the irradiance of a small bulb at different distances using a light meter in a darkened room in Maynooth. | Distance $d$ / m | Irradiance $I$ /…

A student measures the irradiance of a small bulb at different distances using a light meter in a darkened room in Maynooth.

Distance $d$ / m Irradiance $I$ / $\text{W m}^{-2}$
0.50 8.0
1.00 2.0
2.00 0.50
  1. a) Show that the data are consistent with an inverse-square law.
  2. b) Calculate the power of the bulb that is emitted as radiation, assuming it emits uniformly in all directions.
  3. c) State one reason why the real value might differ from this calculation.
Solve

Sunlight, angle and focusing

The Sun is a point source far away. Its power is $3.8\times10^{26}\ \text{W}$ and the Earth is $1.5\times10^{11}\ \text{m}$ away, so at the top of the atmosphere

$$I = \frac{3.8\times10^{26}}{4\pi(1.5\times10^{11})^2} \approx 1.4\times10^{3}\ \text{W m}^{-2}.$$

This is the solar constant, about $1{,}400\ \text{W m}^{-2}$.

The Earth intercepts sunlight over a circle of area $\pi R^2$ (a disc), but its surface is a sphere of area $4\pi R^2$. So the average over the whole surface is a quarter of the solar constant, because the same intercepted power is spread over a surface four times larger (this accounts for night and slanted areas).

Tilted surfaces

If rays strike a surface at an angle $\theta$ to the normal, the same beam is spread over a larger area, and the effective irradiance is

$$I_{eff} = I\cos\theta.$$ Parallel rays hitting a tilted surface: the surface sees only the component of the irradiance along its normal, I cos θ. normal surface θ

This is why Irish panels face south and are tilted, and why winter sunlight (low in the sky) gives less than summer sunlight. Cloud and atmosphere also absorb and scatter, so the irradiance on the ground in Ireland is below the solar constant.

Focusing

A lens gathers the power from a wide beam onto a tiny spot. The power is the same but $A$ shrinks, so $I$ grows enormously. Since area scales with the square of the diameter, reducing a beam's diameter by a factor of 100 raises irradiance by $100^2 = 10^4$. The lens of the eye does exactly this to light entering the pupil, focusing it onto a tiny patch of the retina. That is why even a low-power laser can damage an eye.

Practice★★★Not started
The Sun has a radiated power of $3.8\times10^{26}\ \text{W}$. The mean Earth–Sun distance is $1.5\times10^{11}\ \text{m}$ and the radius of the Earth is…

The Sun has a radiated power of $3.8\times10^{26}\ \text{W}$. The mean Earth–Sun distance is $1.5\times10^{11}\ \text{m}$ and the radius of the Earth is $6.4\times10^{6}\ \text{m}$.

  1. a) Calculate the irradiance of sunlight at the top of the Earth's atmosphere (the solar constant).
  2. b) Calculate the total power intercepted by the Earth, treating it as a disc of radius $6.4\times10^{6}\ \text{m}$.
  3. c) Explain why the average irradiance over the Earth's whole surface is one quarter of the value in part a).
  4. d) Explain why the irradiance on the ground in Ireland is lower than the value in part a), giving at least two reasons.
Solve
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A flat solar water heater of area $3.0\ \text{m}^2$ is tilted so that sunlight of irradiance $900\ \text{W m}^{-2}$ strikes it at an angle of $40^\circ$ to the normal of…

A flat solar water heater of area $3.0\ \text{m}^2$ is tilted so that sunlight of irradiance $900\ \text{W m}^{-2}$ strikes it at an angle of $40^\circ$ to the normal of the panel. The panel absorbs $85\%$ of the incident energy, and all of it heats water. Water flows through at a rate of $0.020\ \text{kg s}^{-1}$ with specific heat capacity $4,180\ \text{J kg}^{-1}\text{K}^{-1}$.

  1. a) Explain why the effective irradiance on the panel is $I\cos\theta$, and calculate it.
  2. b) Calculate the power absorbed by the panel.
  3. c) Calculate the temperature rise of the water as it flows through the panel.
  4. d) State the effect on the temperature rise of doubling the flow rate, and justify your answer.
Solve
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A $5.0\ \text{mW}$ laser beam has a circular cross-section of diameter $2.0\ \text{mm}$. It is focused by a lens to a spot of diameter $20\ \mu\text{m}$. a) Calculate…

A $5.0\ \text{mW}$ laser beam has a circular cross-section of diameter $2.0\ \text{mm}$. It is focused by a lens to a spot of diameter $20\ \mu\text{m}$.

  1. a) Calculate the irradiance in the original beam.
  2. b) Calculate the irradiance at the focused spot, assuming all the power passes through the lens.
  3. c) Calculate the ratio of the two irradiances and explain the result.
  4. d) Explain why looking into even a low-power laser beam can damage the eye, with reference to the focusing action of the eye's lens.
Solve

Sources across the electromagnetic spectrum

Irradiance is measured for any electromagnetic (EM) radiation, not only light. All EM waves travel at $c = 3.0\times10^{8}\ \text{m s}^{-1}$ in a vacuum and obey $c = f\lambda$. Each region has typical sources and uses.

Region Typical source Use or effect
Radio waves Aerials, transmitters Broadcasting, Wi-Fi
Microwaves Magnetron in a microwave oven Heating food, mobile phones, radar
Infrared Warm objects, remote controls Heating, thermal imaging
Visible Sun, bulbs, LEDs Vision, photosynthesis
Ultraviolet Sun, tanning lamps Tanning, sterilising; can damage skin
X-rays X-ray tube Medical and dental imaging
Gamma rays Radioactive nuclei Cancer treatment, sterilising

Moving from radio to gamma, frequency and photon energy increase and wavelength decreases. The highest-energy photons are in the gamma region. Ultraviolet, X-rays and gamma rays are ionising: their photons carry enough energy to remove electrons from atoms and can damage living cells.

The electromagnetic spectrum from radio waves to gamma rays; wavelength decreases and frequency increases to the right. Radio Microw IR Visible UV X-ray Gamma Longer λ, lower f →→→ shorter λ, higher f

Vuelve al principio: a panel in Cork receives a lower irradiance than one in Athens because the Sun is lower in the sky (a larger angle $\theta$ and a longer path through the atmosphere), and there is more cloud. Same panel area, less $P = IA$.

Practice★☆☆Not started
Name the region of the electromagnetic spectrum associated with each of the following sources. a) A microwave oven in a kitchen in Drogheda. b) A TV remote control. c) A…

Name the region of the electromagnetic spectrum associated with each of the following sources.

  1. a) A microwave oven in a kitchen in Drogheda.
  2. b) A TV remote control.
  3. c) A dental X-ray machine.
  4. d) A tanning lamp.
  5. e) Which of these has photons of the greatest energy?
Solve

Summary

  • Irradiance is power per unit area: $I = P/A$, in $\text{W m}^{-2}$. Then $P = IA$ and $E = Pt$ (use kWh for household energy).

  • Useful output is efficiency × input power, so a solar farm produces $\text{efficiency} \times IA$.

  • For a point source, $I = P/4\pi r^2$: irradiance follows an inverse square law, $I \propto 1/r^2$.

  • On a tilted surface the effective irradiance is $I\cos\theta$, with $\theta$ measured from the normal.

  • Focusing reduces area and so raises irradiance greatly, which is why laser beams can harm the eye.

  • Across the EM spectrum from radio to gamma, frequency and photon energy rise and wavelength falls; UV, X-rays and gamma rays are ionising.

See all the exercises in this section