Minimum Detectable Wavelength Of A Wire Mesh Radio Telescope?

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Retsek
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The original dish design of the Lovell Radio Telescope at Jodrell Bank used a 50 mm open wire mesh. Estimate the minimum wavelength detectable using this design.

The mark scheme says lambda/20 = 0.05, lambda = 1.

Where the hell did they pluck this 20 from?
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HedgePig
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50mm = 1/20th of 1 meter, the standard SI unit.)

(Disclaimer: my knowledge of astronomy is limited to being able to identify correctly the moon at night. And I’m a big fan of Thomas Dolby’s “The Golden Age of Wireless”)
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Retsek
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(Original post by HedgePig)
50mm = 1/20th of 1 meter, the standard SI unit.)

(Disclaimer: my knowledge of astronomy is limited to being able to identify correctly the moon at night. And I’m a big fan of Thomas Dolby’s “The Golden Age of Wireless”)
Also 50mm = 0.05 m ( The other half of the equation)
So assuming what you said is correct than no matter what the separation of the mesh is, the minimum wavelength would be 1 meter?

e.g. 10mm separation, lambda/10 = 0.01, lambda = 1m
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User1739493574
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Landa/20 is the minimum to avoid spherical aberration
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User1739493574
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This is your minimum angular resolution the 20 does not change
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User1739493574
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So 0.05*20=1m wavelength
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User1739493574
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(Original post by Retsek)
The original dish design of the Lovell Radio Telescope at Jodrell Bank used a 50 mm open wire mesh. Estimate the minimum wavelength detectable using this design.

The mark scheme says lambda/20 = 0.05, lambda = 1.

Where the hell did they pluck this 20 from?
Lambda/20 =minimum to avoid spherical aberration lambda/20 is just given in the cgp textbooks
So 50 mm wire mesh
0.05*20=1m
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