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

[burgerkin]

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

\theta = 1.22 (\lambda / D

The Attempt at a Solution

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

for part b and c, I don't know how to solve. Please help.

 

[burgerkin]

bump bump

 

[gneill]

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.

 

[burgerkin]

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 \theta

then dia = 2r, but the answer is wrong

part c I used d =L \theta

wrong answer again, I guess cause my value for part a is wrong

 

[burgerkin]

Is the telescope a circular aperture? If not, I do not need 1.22 factor in that equation.

 

[gneill]

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?

 

[burgerkin]

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 \mum And again, it was wrong..

 

[burgerkin]

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.

 

[burgerkin]

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.

 

[gneill]

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.