Resolving Power of Telescopes (AQA A Level Physics): Revision Note

The terms 'resolution' and 'resolving power' are often both used interchangeably to describe the quality of a telescope in terms of the minimum angular separation it can achieve

For example, saying

• A telescope has a resolution (or resolving power) of 0.005 degrees

Is the same as saying

• A telescope can resolve two stars which have an angular separation of at least 0.005 degrees

Remember that the 'Airy disc' is just the central maximum of the interference pattern, not the name of the whole pattern itself.

The supermassive black hole at the centre of the Milky Way galaxy is 25 000 light years from Earth. It has a Schwarzschild radius of 1.2 × 10¹⁰ m and emits radio waves at a frequency of 230 GHz.

The Event Horizon Telescope (EHT) is an array of radio telescopes which has the same resolution as a single radio telescope with a diameter of 8000 km.

(a) Calculate the minimum angular separation which could be resolved by the EHT. [2]

 (b) Deduce whether the resolution of the EHT is suitable to obtain a detailed view of the supermassive black hole. Support your answer with a calculation. [3]

 Answer:

Part (a) 

Step 1: List the relevant quantities

• Frequency of the radio waves, $f=230\mathrm{GHz}=230\times {10}^9\mathrm{Hz}$

• Speed of light, $c=3\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}$

• Diameter of EHT, $D=8000\mathrm{km}=8000\times {10}^3\mathrm{m}$

Step 2: Write down the equation for the minimum angular separation and substitute in the wave equation 

• Wave equation: $\lambda =\frac{c}{f}$

• Minimum angular separation:  $\theta =\frac{\lambda }{D}=\frac{c}{fD}$ [1 mark]

Step 3: Calculate the minimum angular separation the EHT can resolve

$\theta =\frac{3\times {10}^8}{\left(230\times {10}^9\right)\times \left(8000\times {10}^3\right)}=1.63\times {10}^{-10}\mathrm{rad}$ [1 mark]

• This means the EHT can obtain a detailed view of objects down to an angular size of 1.63 × 10⁻¹⁰ rad

Part (b)

Step 1: List the relevant quantities

• Diameter of the black hole, $s=2\times \left(1.2\times {10}^{10}\right)=2.4\times {10}^{10}\mathrm{m}$

• Light year, $1\mathrm{ly}=9.46\times {10}^{15}\mathrm{m}$ (included in the data booklet)

• Distance to black hole, $d=25000\mathrm{ly}$

Step 2: Write down the equation for angular size

• Angular size = angle subtended by the black hole (i.e. 2 × Schwarzschild radius): 

$s=d\theta \Rightarrow \theta =\frac{s}{d}$

Step 3: Carry out a supporting calculation and write a conclusion

You could use...

Method 1:  Calculate the angle subtended by the black hole

• At a distance of 25 000 ly, the black hole subtends an angular size of:

$\theta =\frac{s}{d}$

$\theta =\frac{2.4\times {10}^{10}}{25000\times \left(9.46\times {10}^{15}\right)}=1.01\times {10}^{-10}\mathrm{rad}$ [1 mark]

• Compare with the resolution of the EHT:

So, $1.63\times {10}^{-10}\mathrm{rad}>1.01\times {10}^{-10}\mathrm{rad}$

• Conclusion:

  • The angle subtended by the black hole, 1.01 × 10⁻¹⁰ rad, is approximately 1.6 times greater than the limit of the resolution of the EHT, 1.63 × 10⁻¹⁰ rad

  • Therefore, it will not be able to suitably resolve detail in an image of the black hole [1 mark]

  • The angle of the black hole as viewed from Earth is wider than the angle that the telescope can view [1 mark]

Or, you could use...

Method 2:  Calculate the size of an object that could just be resolved at this distance

• At a distance of 25 000 ly, the smallest object the EHT could resolve in detail is:

$s=d\theta$

$s=25000\times \left(9.46\times {10}^{15}\right)\times \left(1.63\times {10}^{-10}\right)=3.85\times {10}^{10}\mathrm{m}$ [1 mark]

• Compare with the diameter of the black hole:

So, $3.85\times {10}^{10}\mathrm{m}>2.40\times {10}^{10}\mathrm{m}$

• Conclusion: 

  • The smallest object the EHT could just resolve at this distance, 3.85 × 10¹⁰ m, is approximately 1.6 times greater than the diameter of the black hole, 2.40 × 10¹⁰ m

  • Therefore, it will not be able to suitably resolve detail in an image of the black hole [1 mark]

  • The distance between two objects that can be identified as two when viewed from Earth is much greater than the diameter of the black hole [1 mark]

It is better to say that $\theta$ is the 'minimum angular resolution' of the telescope instead of 'resolving power', as this implies that $\theta$ is a power (in watts) instead of an angle. However, if you are asked for the resolving power in the exam, it means to calculate $\theta$.

Remember that the wavelength and diameter must be in the same units.