Astronomy - Ch. 6: Telescopes (7 of 25) Finding Resolutions of Craters on the Moon
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- čas přidán 14. 10. 2024
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In this video I will calculate the telescope's resolutions of craters on the Moon.
Such great videos. I had never received so much knowledge in my life in such short time before. I don't wanna sleep I just wanna keep watching.
Wonderful!
Sir thank you for the very informative video....I would like to ask a question though....how is resolution calculated in realistic world where light is not a single wavelength.. thank you in advance
You are correct is assuming that light is made up of different wavelengths, such that the resolution is not going to be an exact number. We usually take the most common wavelength (the peak of the blackbody radiation curve) as an "average" wavelength.
Very nice explanation….
Thank you. Glad you liked it. 🙂
can you explain how the 2.5e5 coefficient was derived?
It is actually 2.512^5 which is equal to 100 They wanted 5 magnitudes to be equated to an increase in intensity of 100
Why is the coefficient before lambda divided by diameter 250,000? How is this equation derived
The angle of resolution for a circular aperture is: theta = 1.22 lambda/ D but that gives the answer in radians. In astronomy it makes more sense to adapt the constant so we get the anwer in arc seconds.
Sir, I was reading about resolving power and I found that the constant was 1.22 somewhere it was 1.1 and in this video you used 2.5*10^5 can you please guide me? Thank you in advance.
The formula calculates the angle in arcseconds. I guess 1.22 is giving radians as answer (converting 2.5*10^5 arcseconds into radians gives 1,22 radians)
Great Videos
Sir , when I calculate 2.5x10^5 ( 500x10^-9 / 0.1 ) I get 1.25 not 125 . What am I missing ? Thanks in advance !
you are correct. I probably missed the decimal point on my calculator. (eyes aren't what they used to be)
Michel van Biezen Haha . No worries . Thanks , professor !
@@MichelvanBiezen you are cool professor......I am your big fan from bangladesh
I don't understand what 2.5e5 means, is that a constant of the equation? something about the telescope? I would like to know what it is.
It is a constant particular to how lenses and reflecting mirrors work in telescopes. See this video for more information: Physics - Diffraction of Light (4 of 4) Circular Diffraction Patterns
thanks for the answer...
And how fast you did it.
Very tiny D :DDD. Great video
For 5 m diameter one won't it it 0.025 arc second
Yes it will. Someone else already pointed out the error in the first calculation. My guess is something similar happened in the second... I'm not sure. I realized that the second could not possibly be 2.5 arcseconds as that's a bigger resolution angle than the first calculation after it was corrected. At least now we have the equation though.
Note that in the first equation the wavelength is entered in meters and in the second equation the wavelength is entered in micro meters.
@@MichelvanBiezen ahh ok thank you for clarity sir!
6:13 sir iam getting 0.025 😟
Yes, you are correct. (I probably pushed the wrong buttons)
@@MichelvanBiezen oh thank you sir 😊 and have a nice day
you are awesome