Calculating Slit Width from Diffraction Pattern

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In summary, the diffraction pattern produced by a He-Ne laser passing through a single slit with a wavelength of 632.8 nm and viewed on a screen 1.58 m behind the slit can be used to find the slit width using the equation a = {(wavelength)D}/(distance of minima). However, the distance of the minima must be measured from the center of the pattern or multiplied by 'm' for accurate results.
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Octoshark
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Homework Statement



The diffraction pattern shown in the figure is produced by passing He-Ne laser light (λ = 632.8 nm) through a single slit and viewing the pattern on a screen 1.58 m behind the slit.

What is the slit width?

28-38.gif


Homework Equations



a = {(wavelength)D}/(distance of minima)

The Attempt at a Solution



a = {(632.8 x 10^-9)(1.58)}/(.152)

Answer: slit width = 6.577 x 10^-6 which is very wrong.

I feel like my equation is definitely wrong.
 
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  • #3
So by dividing by .076 instead of .152 I get 1.316 x 10^-5 which is incorrect. I must be misunderstanding you, sorry.
 
  • #4
Octoshark said:
So by dividing by .076 instead of .152 I get 1.316 x 10^-5 which is incorrect.
0.076 would be the distance from the center to which minima?
 
  • #5
Ahhh, got it. Thanks a bunch ;)
 

Related to Calculating Slit Width from Diffraction Pattern

What is a basic diffraction problem?

A basic diffraction problem involves studying the behavior of waves, such as light or sound, as they pass through an obstacle or aperture. It is used to understand how these waves interact with other objects and can help predict patterns and behavior.

What is the difference between diffraction and refraction?

Diffraction is the bending and spreading of waves as they pass through an obstacle or aperture, while refraction is the change in direction of waves as they pass through a medium with a different density. Diffraction does not involve a change in medium, while refraction does.

What is the relationship between wavelength and diffraction?

The wavelength of a wave determines the amount of diffraction that will occur. The smaller the wavelength, the less the wave will diffract. This is why we can see visible light but not radio waves, as the wavelength of visible light is smaller and can diffract around obstacles more easily.

How is the intensity of a diffracted wave affected by the size of the obstacle or aperture?

The intensity of a diffracted wave is directly proportional to the size of the obstacle or aperture. This means that a larger obstacle or aperture will result in a more intense diffracted wave.

What are some real-life applications of the basic diffraction problem?

Diffracted waves are used in many technologies, such as radio antennas, microscopes, and telescopes. Diffraction is also important in understanding the behavior of light in photography and holography. In addition, diffraction is studied in fields such as seismology and acoustics to better understand the behavior of waves in different environments.

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