What is the ratio of the largest slit width to the smallest?

  • #1
I'm not exactly sure how to get started on this one. Can anybody help me?...Please?

Two different single slits are used in an experiment involving one source of monochromatic light. With slit 1 in place, the first dark fringe is observed at an angle of 45 degrees. With slit 2, the first dark fringe is observed at an angle of 55 degrees.

a) Which slit is widest? Why?

b) What is the ratio of the largest slit width to the smallest?
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
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You need to show some work before we can help you. Do you know any equations to do with diffraction through a slit?
 
  • #4
The equation I've tried was sintheta = lamda/w

For a) I plugged in an arbitrary wavelength of 720 nm and found that slit ! was larger and I figure its because the first dark fringe is observed at less of an angle than slit 2. Therefore, the number of pairs of rays that destructively interfere are increased which causes the maxima to decrease.

For b) I used the arbitrary wavelength of 720 nm and plugged it in for both cases (45 degrees and 55 degrees) which resulted in a ratio of 1.17:1

Does this sound plausible to you guys?
 
  • #5
Doc Al
Mentor
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The equation I've tried was sintheta = lamda/w
That gives the condition for the first minima. Since you're interested in the width, you can rewrite it as:
[tex]w = \lambda/\sin \theta[/tex]
For a) I plugged in an arbitrary wavelength of 720 nm and found that slit ! was larger and I figure its because the first dark fringe is observed at less of an angle than slit 2. Therefore, the number of pairs of rays that destructively interfere are increased which causes the maxima to decrease.
Since for angles between 0 and 90 degrees sin(theta) increases with angle, that equation should tell you that a bigger angle implies a narrower slit.

For b) I used the arbitrary wavelength of 720 nm and plugged it in for both cases (45 degrees and 55 degrees) which resulted in a ratio of 1.17:1
Close enough. But no need to plug in a particular value of wavelength. (Although that's perfectly OK.) Try using the equation to relate a ratio of slit widths to a ratio of sines of the angles.
 
  • #6
Thanks Doc! You've been a big help and I think I now know what to do. Cheers!
 

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