A person uses a re-flecting telescope of dia= 20 cm

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In summary: Cheers.In summary, a reflecting telescope with a diameter of 20 cm and a focal length of 170 cm can achieve a minimum angular resolution of 0.00021 degrees (or 0.755 arcseconds) for observing light with a wavelength of 600 nm. The diameter of the Airy disk for this setup would be 714 micrometers. The minimum separation distance of two objects on Mars that the telescope can resolve would depend on their angular separation on Earth, which is equal to the angular Airy radius of the telescope. However, it appears that for this problem, the approximation without the 1.22 factor is used instead.
  • #1
burgerkin
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Homework Statement




A person uses a re-flecting telescope of dia= 20 cm and f=170 cm to observe light of λ = 600 nm from a star.
(a) find minimum angular resolution can he get?
°

(b) find the diameter of the Airy disk?
µm

(c) find the minimum separation distance of two objects on the Mars that the telescope can resolve? (The Earth-Mars distance is 3.83 10^8 m.)


Homework Equations



[tex]\theta[/tex] = 1.22 ([tex]\lambda[/tex] / D




The Attempt at a Solution



for part A, I used the relation above to solve for [tex]\theta[/tex] = 0.00021 degree

for part b and c, I don't know how to solve. Please help.
 
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  • #2
bump bump
 
  • #3
You've calculated the angular radius of the Airy disk as θ = 0.00021 degrees (arcseconds would be a nice unit to use here), which corresponds to the angular resolving power of the telescope.

Use θ and the focal length to determine the linear diameter of the Airy disk. For small angles, the linear projection at distance r is rθ. Remember, θ corresponds to the radius, not the diameter.

For part c, determine what the linear separation of two objects has to be at the distance of Mars in order for the angular separation as observed on Earth to be equal to the angular Airy radius of the scope.
 
  • #4
Thanks so much!

However, I got wrong answers for all 3 parts. I followed your instructions, so I am thinking I might have done the first one wrong.

Do I need the factor of 1.22 for the first formula? Since it says diameter, I used the equation with 1.22.

For part b, I let p(object distance goes to infinity) to get q, the image distance, so I got q=1.7 m, than r = 1.7 x [tex]\theta[/tex]

then dia = 2r, but the answer is wrong

part c I used d =L [tex]\theta[/tex]

wrong answer again, I guess cause my value for part a is wrong
 
  • #5
Is the telescope a circular aperture? If not, I do not need 1.22 factor in that equation.
 
  • #6
For part (a), the formula for the angular radius of the Airy disk is sin(θ) = 1.22λ/D as you've stated. For the given data you obtained a result of θ = 0.000021°. I might have used scientific notation here, and said θ = 2.10 x 10-4 °, or converted to arcseconds,
θ = 0.755''.

For part (b) it seems that they want the linear diameter of the Airy disk in micrometers (µm).
You've got the angular radius θ and the focal length d = 170cm. As you stated, the diameter should be 2*θ*d using the small angle approximation. Did you remember to convert your angle to radians? can you show your calculation and result?
 
  • #7
For part a, I was asked to give the answer in degree, so I did not convert to radians. So the calculation did give me 0.00021 degree, but it was wrong answer.

For part b, I did 2 x 1.7 x 0.00021 =7.14 e-4, and I did convert into micro meter, 714 [tex]\mu[/tex]m And again, it was wrong..
 
  • #8
I just realized that I did not use radians for part b.

Anyhow, 0.00021 degree is a wrong answer. But the question is asking angle in degrees.
 
Last edited:
  • #9
Hi Thanks a lot!

I just found out that for telescope, we use the equation without 1.22 factor. And I tried. answers are all correct.
 
  • #10
burgerkin said:
Hi Thanks a lot!

I just found out that for telescope, we use the equation without 1.22 factor. And I tried. answers are all correct.

Hmm. That's a pretty common approximation I suppose; The "true" formula should include the 1.22 constant. I'm a bit surprised that the system accepts a "crude" approximation and rejects the more accurate result. Ah well, I'm glad that you succeeded.
 

Related to A person uses a re-flecting telescope of dia= 20 cm

1. How does a reflecting telescope work?

A reflecting telescope works by using a curved mirror to collect and reflect light, which is then focused onto a smaller secondary mirror. This secondary mirror reflects the light out the side of the telescope and into the eyepiece, allowing you to view distant objects.

2. What is the advantage of using a reflecting telescope over a refracting telescope?

The main advantage of a reflecting telescope is that it does not suffer from chromatic aberration, which is a distortion of color that can occur in refracting telescopes due to the bending of light through different mediums. Reflecting telescopes also tend to be less expensive and easier to manufacture than refracting telescopes.

3. How does the diameter of the telescope affect its performance?

The diameter, or aperture, of a telescope is directly related to its light-gathering ability. A larger aperture allows more light to enter the telescope, resulting in a brighter and clearer image. This is why a telescope with a diameter of 20 cm will generally provide better views than one with a smaller diameter.

4. Can a reflecting telescope be used for both terrestrial and astronomical viewing?

While reflecting telescopes are primarily designed for astronomical viewing, they can also be used for terrestrial viewing. However, they may not be as suitable for this purpose as refracting telescopes, which typically have a wider field of view and can provide sharper images for terrestrial objects.

5. Is a 20 cm telescope considered large?

In the world of amateur astronomy, a 20 cm telescope would generally be considered a medium-sized telescope. However, it is still large enough to provide impressive views of celestial objects and is a popular size for serious amateur astronomers.

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