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

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Homework Help Overview

The problem involves analyzing a single slit diffraction experiment with two different slit widths and their corresponding angles for the first dark fringe. The participants are tasked with determining which slit is wider and calculating the ratio of the largest slit width to the smallest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between slit width and the angle of the first dark fringe, referencing the diffraction pattern and relevant equations. Questions about the application of the diffraction equation and the implications of angles on slit width are raised.

Discussion Status

Some participants have provided guidance on using the diffraction equation and have explored the implications of angles on slit width. There is an ongoing exploration of the relationship between the slit widths and the angles observed, with some participants suggesting a method to relate the ratio of slit widths to the sine of the angles.

Contextual Notes

Participants are encouraged to show their work and consider the implications of using arbitrary values for wavelength in their calculations. The discussion acknowledges the need for clarity on the underlying principles of diffraction.

Mitchtwitchita
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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?
 
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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?
 
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?
 
Mitchtwitchita said:
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:
w = \lambda/\sin \theta
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.
 
Thanks Doc! You've been a big help and I think I now know what to do. Cheers!
 

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