Luminosity and intensity of starsEdexcel A-Level Physics: Revision notes

Section 1

Luminosity

The luminosity $L$ of a star is the total power it radiates across the whole electromagnetic spectrum. It is measured in watts (W). The Sun's luminosity is $3.85 \times 10^{26}$ W.

Luminosity is a property of the star itself. It does not change with how far away you are.

Key termsluminosity

Common mistake

Do not confuse luminosity with how bright a star looks. Apparent brightness also depends on distance.

Section 2

Intensity

The intensity $I$ of radiation is the power per unit area received at right angles to the direction of travel. Its unit is W m⁻².

Intensity depends on the luminosity of the source and on the distance from it. A dim, nearby star can give more intensity at the Earth than a very luminous but distant one.

Key termsintensity

Section 3

The inverse square law: I = L/4πd²

Suppose a star radiates equally in all directions. At distance $d$ the power is spread uniformly over a sphere of surface area $4\pi d^2$ , so

$I = \dfrac{L}{4\pi d^2}$

Intensity is therefore proportional to $1/d^2$ . Doubling the distance reduces the intensity to one quarter, and tripling it reduces it to one ninth.

The equation assumes the star radiates equally in all directions and that no light is absorbed or scattered on its way.

Key termsinverse square law

Common mistake

Do not forget to square d. The area of the sphere is 4πd², so the intensity falls with the square of the distance.

Section 4

Worked examples

Intensity at the Earth. The Sun has $L = 3.85 \times 10^{26}$ W and $d = 1.5 \times 10^{11}$ m. Then $I = 3.85 \times 10^{26} / (4\pi \times (1.5 \times 10^{11})^2) = 1.4 \times 10^{3}$ W m⁻².

Luminosity from a measured intensity. A star at $8.1 \times 10^{16}$ m gives $1.2 \times 10^{-7}$ W m⁻². Then $L = 4\pi d^2 I = 4\pi \times (8.1 \times 10^{16})^2 \times 1.2 \times 10^{-7} = 9.9 \times 10^{27}$ W.

Distance. Rearranging gives $d = \sqrt{L / 4\pi I}$ .

Key termsrearranging I = L/4πd²

Section 5

Comparing stars

Two sources with the same luminosity give the same intensity only if they are the same distance away. For stars of equal intensity, $L_1/d_1^2 = L_2/d_2^2$ . A star with four times the luminosity of another gives the same intensity at twice the distance.

If two stars are at the same distance, the ratio of intensities equals the ratio of luminosities.

Must Know

Luminosity: total power radiated, in W
Intensity: power per unit area, in W m⁻²
$I = L/4\pi d^2$ , an inverse square law
Assumptions: radiation emitted equally in all directions, and no absorption on the way

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Luminosity and intensity of stars

An astronomer measures the intensity of the light received from distant stars using a telescope fitted with a sensitive detector.
Explain why the intensity of the light received from a star decreases as the distance from the star increases.2 marks
The Sun has a luminosity of 3.85 × 10²⁶ W. The mean distance from the Sun to the Earth is 1.5 × 10¹¹ m. Assume that the Sun radiates equally in all directions and that no radiation is absorbed on its way to the planets.
The planet Mars is 2.3 × 10¹¹ m from the Sun. Calculate the intensity of sunlight at Mars.2 marks
Astronomers measure the intensity of light from a bright star, S, as 1.2 × 10⁻⁷ W m⁻² at the Earth. Star S is 8.1 × 10¹⁶ m from the Earth.
Calculate the luminosity of star S.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).