These problems relate to material to be covered in class on Tuesday, Nov. 9th.
I have written comments on several of the problems in the instructions that you will see when you look at the assignment. Please read them before you work on the assignment.
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6 comments:
Prof Stephens,
I am having some difficulty on problem 8 part c. For my question it states "if a telescope is to be constructed with a tube 10cm long and a magnification of 3.80, what are the focal length of the objective and eyepiece."
I started to tackle the problem by equating magnification to the focal length of the objective and the eyepiece using the formula
m=(h'/f_e)/(h'/f_o)=> m=f_o/f_e
also since the tube length is equal to f_o+f_e then i can substitute for one and solve for the other. I set that 10cm=f_o+f_e then f_e=10-f_o
into the original equation 3.8=f_o/(10-f_o) solving for f_o=7.916cm and for f_e=2.083cm
However I know that this process is incorrect and any assistance that you can provide me with is greatly appreciated.
Problem 8 is about the Galilean telescope, which has a diverging (negative focal length) eyepiece. So your basic approach is correct, but f_e should be negative. How does that change the equations? S&V are a bit sloppy about the sign of the magnification of a telescope, since an astronomical telescope (both focal lengths positive) produces an inverted image.
I know I'm posting this really late, but I just now had a chance to finish number 12.
I am using the equation (n-1)=Nλ/2L (N=170, L=.05m, λ=560*10^-9m) and when I submit my answer it tells me I'm wrong.
I am getting a number that looks like this: 1.000### Webassign says to submit it to 5 decimal points, so I rounded and typed in 1.000## and it was wrong. I thought it maybe counted the 1 as a decimal place so I typed in 1.000# and still it was wrong. I have tried recalculating several times and I continue to get the answer wrong. What could my problem be?
Look at the last slide of yesterday's lecture and you'll see what's wrong with your formula. Recall (SV8, 3rd full paragraph on page 840) that "the term fringe shift is used to describe the change in a fringe from dark to light or from light to dark."
I read the paragraph and I'm sorry to say that the wording used by this author only serves to confuse me more than I already am. (I rarely use the text for more than looking up equations and constants and it's occasional use as a paperweight)
Anyway, I see that on the slide you have Nshifts=[(L/λn)-(L/λ)] but what is λn?? I had this formula written down when I began the problem, but I have no idea what λn is so I wasn't sure if that was what I needed to use. You also have down that N=λ/4..which do I use and how?
Nshifts = 4 (L/lambdan - L/lambda), where lambda is the wavelength in vacuum and lambdan is the wavelength in the medium of index n (air, in this case). So lambdan = lambda/n, and Nshifts = 4*L*(n-1)/lambda.
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