Destructive interference through a narrow slit

In summary, the conversation discusses the calculation of width using the equation angle=wavelength/width of slit and the incorporation of a constant of 28.7. It is discovered that the constant is half of the coefficient between degree and radian, and this leads to the realization that the angle value of degree was used instead of radian. When converting to radians, the resulting answer needs to be divided by 2 for it to be correct. The equation also highlights the difference between enhanced and destructive interference.
  • #1
JoeyBob
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Homework Statement
See attached
Relevant Equations
angle=wavelength/a
So I thought angle=wavelength/width of slit

But when I solve for the width I got the wrong answer of 4567 nm, when the answer is suppose to be 130881 nm. I figured out that I needed to multiply my incorrect answer by 28.7, but where does this constant come from? Its not part of the equation when there's constructive interference.
 

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  • #2
It might be half of coefficient between degree and radian
[tex]360 / 2\pi = 57.29..[/tex]
 
  • #3
anuttarasammyak said:
It might be half of coefficient between degree and radian
[tex]360 / 2\pi = 57.29..[/tex]
How does that convert to multiplying wavelength/angle by 28.7?
 
  • #4
This coincidence hinted me that you might have used angle value of degree not radian in the calculation. How did you do it?
 
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  • #5
anuttarasammyak said:
This coincidence hinted me that you might have used angle value of degree not radian in the calculation. How did you do it?

Youre right, not used to seeing degrees so small. But when I convert it to radians I now have to divide the answer I get by 2 to get the right answer, why is that?

width = wavelength/angle = 540 nm /0.00209 radians, but this gives me an answer that needs to be divided by 2 to get the right answer.
 
  • #6
For enhanced interference
[tex]d\sin\theta=\lambda[/tex]
For destructive interference
[tex]d\sin\theta=\frac{\lambda}{2}[/tex]
, a half of the former.
 
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Related to Destructive interference through a narrow slit

1. What is destructive interference through a narrow slit?

Destructive interference through a narrow slit is a phenomenon that occurs when two or more waves with the same frequency and amplitude meet at a narrow opening, resulting in a cancellation of the waves' amplitudes at certain points.

2. How does destructive interference through a narrow slit occur?

When waves pass through a narrow slit, they diffract and spread out, creating a pattern of alternating high and low intensity regions. If two or more waves meet at a point where the peaks of one wave align with the troughs of another, they will cancel each other out, resulting in a lower overall intensity at that point.

3. What factors affect the degree of destructive interference through a narrow slit?

The degree of destructive interference through a narrow slit is affected by the wavelength of the waves, the width of the slit, and the distance between the slit and the point of observation. In general, shorter wavelengths, narrower slits, and smaller distances between the slit and the observation point will result in a more pronounced interference pattern.

4. What are some real-world applications of destructive interference through a narrow slit?

Destructive interference through a narrow slit is used in many everyday technologies, such as radio antennas, diffraction gratings, and acoustic filters. It is also utilized in scientific experiments to study the properties of waves and to create interference patterns for analysis.

5. Can destructive interference through a narrow slit be beneficial?

Yes, destructive interference through a narrow slit can be beneficial in certain situations. For example, it is used in noise-cancelling headphones to cancel out unwanted background noise, resulting in clearer sound for the listener. It is also used in some imaging techniques, such as electron microscopy, to enhance contrast and resolution.

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