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IBPhysicsDPSLRevision NotesWave BehaviourDoppler EffectThe Doppler Effect of Light

The Doppler Effect of Light (DP IB Physics)

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Katie M

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Katie M

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DP Physics: SLDP Physics: HL

The Doppler Effect of Light

  • The Doppler shift for a light-emitting non-relativistic source can be described using the equation:

fraction numerator increment f over denominator f end fraction space equals space fraction numerator increment lambda over denominator lambda end fraction space almost equal to fraction numerator increment v over denominator c end fraction

  • Where:

    • Δf = change in frequency (Hz)

    • f = reference (original) frequency (Hz)

    • Δλ = change in wavelength (m)

    • λ = reference (original) wavelength (m)

    • Δv = relative velocity of the source and observer (m s–1)

    • c = the speed of light (m s–1)

  • The sign ≈ means 'approximately equals to'

  • This equation only works if v << c

  • The change in wavelength Δλ is equal to:

increment lambda space equals space lambda subscript 0 space minus space lambda

  • Where:

    • λ0 = observed wavelength of the source (m)

  • Since the fractions have the same units on the numerator (top number) and denominator (bottom number), the Doppler shift has no units

  • The relative speed between the source and observer along the line joining them is given by:

increment v space equals space v subscript s space minus space v subscript o

  • Where:

    • vs = velocity of the source of the light (m s–1)

    • vo = velocity of the observer (m s–1)

  • Usually, we calculate the speed of the source of electromagnetic waves relative to an observer which we assume to be stationary

    • Therefore vo = 0, hence ∆v = vs = v

    • Where v is the velocity at which the source of the electromagnetic waves is moving from the observer

  • Hence, the Doppler shift equation can be written in terms of wavelength:

fraction numerator increment lambda space over denominator lambda end fraction space equals space fraction numerator lambda subscript 0 space minus space lambda over denominator lambda end fraction space almost equal to space v over c

  • It can also be written in terms of frequency:

fraction numerator increment f space over denominator f end fraction space equals space fraction numerator f subscript 0 space minus space f over denominator f end fraction space almost equal to space v over c

Spectral Lines

25-2-redshift

Spectral lines showing red shift

  • Each line represents an element making up the composition of the galaxy

  • The lines are identical to those measured in the lab and the light measured from the distant galaxy

  • Since the lines all move to the left (the red end of the spectrum) this means the galaxy is travelling away from Earth

Worked Example

A stationary source of light is found to have a spectral line of wavelength 438 nm.  The same line from a distant star that is moving away from us has a wavelength of 608 nm.

Calculate the speed at which the star is travelling away from Earth.

Answer:

Step 1: List the known quantities

  • Unshifted wavelength, λ = 438 nm

  • Shifted wavelength, λ0 = 608 nm

  • Change in wavelength, Δλ = (608 – 438) nm = 170 nm 

  • Speed of light, c = 3.0 × 108 m s–1

Step 2: Write down the Doppler equation and rearrange for velocity v

fraction numerator increment lambda over denominator lambda end fraction space equals space v over c

v space equals space fraction numerator c increment lambda over denominator lambda end fraction

Step 3: Substitute values to calculate v

v space equals space fraction numerator left parenthesis 3.0 blank cross times blank 10 to the power of 8 right parenthesis cross times 170 over denominator 438 end fraction = 1.16 × 108 m s–1

Worked Example

The stars in a distant galaxy can be seen to orbit about a galactic centre. The galaxy can be observed 'edge-on' from the Earth.

Light emitted from a star on the left-hand side of the galaxy is measured to have a wavelength of 656.44 nm. The same spectral line from a star on the right-hand side is measured to have a wavelength of 656.12 nm.

The wavelength of the same spectral line measured on Earth is 656.28 nm.

(a) State and explain which side of the galaxy is moving towards the Earth.

(b) Calculate the rotational speed of the galaxy. 

Answer:

(a)

  • The light from the right-hand side (656.12 nm) is observed to be at a shorter wavelength than the reference line (656.28 nm)

  • Therefore, the right-hand side has been blue-shifted and must be moving towards the Earth

(b)

Step 1: List the known quantities

  • Observed wavelength on LHS, lambda subscript L H S end subscript = 656.44 nm

  • Observed wavelength on RHS, lambda subscript R H S end subscript = 656.12 nm

  • Reference wavelength, λ = 656.28 nm

  • Speed of light, c = 3.0 × 108 m s−1

Step 2: Calculate the average change in wavelength

increment lambda space equals space fraction numerator lambda subscript L H S end subscript space minus space lambda subscript R H S end subscript over denominator 2 end fraction space equals space fraction numerator 656.44 space minus space 656.12 over denominator 2 end fraction

increment lambda space= 0.32 nm

Step 3: Write down the Doppler equation and rearrange for velocity v

fraction numerator increment lambda over denominator lambda end fraction space equals space v over c

v space equals space fraction numerator c increment lambda over denominator lambda end fraction

Step 4: Substitute values into the velocity equation

v space equals space fraction numerator open parentheses 3 cross times 10 to the power of 8 close parentheses cross times 0.32 over denominator 656.28 end fraction

Rotational speed:  v space equals space 1.46 space cross times 10 to the power of 5 space straight m space straight s to the power of negative 1 end exponent space equals space 146 space km space straight s to the power of negative 1 end exponent

Examiner Tips and Tricks

You need to know that in the visible light spectrum red light has the longest wavelength and the smallest frequency compared to blue light which has a shorter wavelength and higher frequency.

The second worked example didn't change the wavelengths from nm into m, since it doesn't matter in the equation as the units will cancel out. 

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.