Having trouble setting up this Trig problem.

  • Thread starter jrjack
  • Start date
  • Tags
    Trig
In summary, the largest refracting telescope in the world is at the University of Chicago, and its resolution is approximately 0.00003754°.
  • #1
jrjack
111
0

Homework Statement


Two stars that are very close may appear to be one. The ability of a telescope to separate their images is called its resolution. The smaller the resolution, the better a telescope's ability to separate images in the sky. In a refracting telescope, resolution θ (see the figure) can be improved by using a lens with a larger diameter D. The relationship between θ in degrees and D in meters is given by sin θ = 1.22λ/D, where λ is the wavelength of light in meters. The largest refracting telescope in the world is at the University of Chicago. At a wavelength of λ = 550 × 10−9 meter, its resolution is approximately 0.00003754°. Approximate the diameter of the lens. (Round your answer to two decimal places.)

Homework Equations



sin(.00003754)=[1.22(550x10^-9)]

The Attempt at a Solution



When I do it set up this way, I get a decimal, which cannot be the size of largest telescope lens.
If I divide like this: sin(.00003754)/1.22(550x10^-9) = 55.9463m this was not right.

I googled the actual telescope lens diameter and it is about 40 ft. (or 12.192m)
 
Physics news on Phys.org
  • #2
Your math skills (and reading comprehension) need work. The lens diameter for the Yerkes telescope is 40 inches (not feet) or 1.02 m

The formula you were given is sin theta = 1.22 lambda / D [Note: the D was omitted from the formula in sec. 2 of the OP]

When you substitute the given data into the given formula, and invert, D = 1.02 m
 
  • #3
hi jrjack! :smile:
jrjack said:
The relationship between θ in degrees and D in meters is given by sin θ = 1.22λ/D…

sin(.00003754)=[1.22(550x10^-9)]

you seem to be using sin θ/1.22λ = D (instead of 1/D) :redface:

(and are you converting to radians?)
 
  • #4
SteamKing said:
Your math skills (and reading comprehension) need work. The lens diameter for the Yerkes telescope is 40 inches (not feet) or 1.02 m

The formula you were given is sin theta = 1.22 lambda / D [Note: the D was omitted from the formula in sec. 2 of the OP]

When you substitute the given data into the given formula, and invert, D = 1.02 m

I do struggle with both skills, but hopefully more practice will help. sorry I forgot to type the / D
The section we are learning has only dealt with right triangles, and has not mentioned Lambda or wavelength yet, so I am having trouble visuallizing how the lens dia is related to theta.
tellescope.gif


I tried it in degrees and radians and I'm still not getting the right answer.
sin (00003754)degrees =.000000655

.000000655/ 1.22 (550x10^-9) = .9764479899 in degrees

sin(.00003754)radians = .0000375399 / 1.22 (550x10^-9) = 55.94634872 in radians

I know I am missing something or not doing something right.
 
  • #5
The argument of the sine function (0.00003754) is already in degrees.

The sine of an angle (whether the angle is measured in degrees or radians) is a pure number without units.

The formula you use to calculate D of the lens is:

sin (theta) = 1.22 lambda / D, where theta is in degrees and lambda and D are in meters

In order to solve for D, you must apply a little algebra to the given formula:

D = 1.22 lambda / sin (theta) [divide both sides by sin (theta) and multiply both sides by D]

Plug your data into the modified formula above to calculate D.
 
  • #6
Thanks.
I'm slowly learning, It's been 14 years since high school, and my algrebra refresher course was lacking.
I now realize when I converted the formula, I ended up with the sin (theta) as the numerator, when it should have been the denominator. Then I had to make sure I put my calculator back in degree mode.
Got it now, thanks.
 

1. What is Trigonometry?

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is used to solve various problems involving triangles and is also used in fields such as engineering, physics, and astronomy.

2. Why do I have trouble setting up Trig problems?

The main reason people have trouble setting up Trig problems is because they lack a strong foundation in the basic concepts and principles of Trigonometry. It is important to understand the definitions, formulas, and rules before attempting to solve more complex problems.

3. How can I improve my Trig problem-solving skills?

Practice is key to improving your Trig problem-solving skills. Make sure to review and understand the concepts before attempting to solve problems. You can also seek help from a tutor or teacher if needed.

4. What are some common mistakes when setting up Trig problems?

Some common mistakes when setting up Trig problems include using the wrong formula, forgetting to convert units, and not drawing accurate diagrams. It is important to double-check your work and be mindful of the units and angles involved in the problem.

5. Can Trigonometry be applied to real-world problems?

Yes, Trigonometry has many real-world applications in fields such as navigation, surveying, architecture, and engineering. It can also be used to solve everyday problems such as finding the height of a building or the distance between two points.

Similar threads

  • Astronomy and Astrophysics
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
796
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
10
Views
3K
Replies
4
Views
1K
Back
Top