Having trouble setting up this Trig problem.

[jrjack]

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)

 

[SteamKing]

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

 

[tiny-tim]

hi jrjack!

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)

(and are you converting to radians?)

 

[jrjack]

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.

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.

 

[SteamKing]

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.

 

[jrjack]

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.