Calculating focal length from a number of focal ratios

  • #1
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Homework Statement


A telescope has a focal ratio of f/7.5. You wish to use it with a spectrometer that requires a f/10 beam as its input. Compute the focal length of a 50mm diameter lens that, when inserted into the beam 150mm in front of the unmodified focal plane, produces the required beam.

Homework Equations


Focal ratio, R = focal length, f / Diameter, D

The Attempt at a Solution


To create a f/10 beam using the 50mm lens, R = f/D so 10 = f / 50, making f = 500mm. This doesn't sound right at all, as it doesn't take into account the telescope's f/7.5 ratio or the fact the lens is 150mm away from the unmodified focal plane. I feel like I'm missing some other equation or important piece of this problem.
 

Answers and Replies

  • #2
andrevdh
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The inserted lens shifts the focal plane.
Also the new focal plane will not lie in the focal plane of the additional lens since the light entering it is not collimated.
 
  • #3
Merlin3189
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I have never heard of this topic before, so any comments are based purely on my understanding of lenses.

My first suggestions are:
Have you drawn a diagram?
What does it mean to say that the spectrometer requires an f/10 beam? Why would this be so?

If my own understanding of the answers to this question is correct, I can suggest how to calculate it.
(Though, other than aligning the optical axes of the lenses (which I assume), I can't see where collimation comes into the calculation.)
 
  • #4
andrevdh
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The cone of light entering the spectrometer needs to be f/10 in order for it to function optimally.
Since this seems to be related to astronomy the light entering the telescope is collimated, not
so for the additional lens that is used to change it to a f/10 beam.
 
  • #5
Merlin3189
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I rather assumed that, "a spectrometer that requires a f/10 beam as its input." meant something like, "The cone of light entering the spectrometer needs to be f/10 in order for it to function optimally." What I was asking was, why or how this is a requirement. What goes wrong if it is greater than f/10 or less than f/10. I thought that might suggest, in conjunction with a diagram, how to achieve it.

I think we may have been talking of different collimations here. Are you just saying that the light from an astronomical object arrives as parallel rays, but the objective focuses them to a converging beam, so that they are no longer parallel when they strike the 50mm lens? Since we are told where we must put the lens, there is nothing we can do to avoid that. My own conclusion was that the purpose of the lens was to correct the convergence of the beam to match the spectroscope's requirement, but not to make the beam again parallel (collimate it?).

I wonder if there is a diagram yet?
 
  • #6
andrevdh
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Yes, that is how I have it too.

I was trying to point the student in the correct direction by making him aware of this situation.

The spectroscope has another collimating lens inside of it.
If the cone of light is not of the right dimensions light is either wasted (it overshoots the collimating lens)
or the collimated beam uses too small an area of the grating or prism to produce an effective spectrum.

The final focal plane is positioned on the entry slit to the spectroscope. The spectrum are then images of the illuminated part of the entry slit.
The spectroscope thus reproduces the astronomical object in slit images in the various parts of the spectrum.

Spectroscopes have to be used correctly since it usually operate with very little light, especially in astronomy.
 
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