- #1
shanepitts
- 84
- 1
Homework Statement
Homework Equations
θ=1.22 (λ/D)
The Attempt at a Solution
Just want to know if I properly answered this question.
Thanks in advance
Did I Properly Answer the Angular Resolution Question?
- Thread starter shanepitts
- Start date
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The discussion confirms that the methodology used to answer the angular resolution question is correct, particularly the application of the formula θ=1.22 (λ/D). However, there is uncertainty regarding the selection of typical wavelengths, as the GMRT operates across a wide frequency range. While 1 m is considered a reasonable ballpark figure, it is noted that radio wavelengths larger than 10 m may be blocked by the atmosphere. The choice of optical wavelength at 500 nm is also acknowledged as a typical value, though it represents a narrower range. Overall, the approach and usage of the formula are deemed appropriate, but clarity on wavelength selection is needed.
θ=1.22 (λ/D)
Just want to know if I properly answered this question.
Thanks in advance
- #2
collinsmark
Science Advisor
Homework Helper
- 3,443
- 3,473
Yes, your methodology is correct.
The only thing I can not confirm is how you are expected to choose the typical wavelengths involved. The GMRT is capable of operating on many frequencies in different bands -- altogether the range is greater than an order of magnitude from highest to lowest wavelength. Although yes, 1 m is within that range, if you're just looking for a ballpark figure.
Similarly, we can say the same thing about the optical wavelength (where 500 nm is typical), although it is a much narrower range (much less than 1 order of magnitude).
So anyway, I'm guessing that your approach is fine. Your usage of the formulae is good. I think your choices of wavelengths are good for ballpark figures, but I can't say for sure that you are supposed to be looking for ballpark figures.
The only thing I can not confirm is how you are expected to choose the typical wavelengths involved. The GMRT is capable of operating on many frequencies in different bands -- altogether the range is greater than an order of magnitude from highest to lowest wavelength. Although yes, 1 m is within that range, if you're just looking for a ballpark figure.
Similarly, we can say the same thing about the optical wavelength (where 500 nm is typical), although it is a much narrower range (much less than 1 order of magnitude).
So anyway, I'm guessing that your approach is fine. Your usage of the formulae is good. I think your choices of wavelengths are good for ballpark figures, but I can't say for sure that you are supposed to be looking for ballpark figures.
- #3
shanepitts
- 84
- 1
Thank you for your detailed responsecollinsmark said:
- #4
paisiello2
- 907
- 88
I think radio wavelengths larger than 10m are blocked by the atmosphere.
The answer is (B) but I don't really understand why.
Based on formula of Young Modulus:
$$x=\frac{FL}{AE}$$
The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A##
I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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