How Does Changing Hole Shapes Affect Diffraction Patterns?

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Remember to always check your calculations and assumptions to ensure accuracy. Good luck with your problem! In summary, the conversation discusses a homework problem involving a laser beam of light passing through a slide with a rectangular pattern of small holes. The small angle approximation is used to determine the positions of the spots on the screen. The reciprocal lattice vectors are used to calculate the diffraction pattern, and S(g) is solved for in both parts of the problem. The second part involves recalculating S(g) with new parameters for square holes. It is important to check calculations and assumptions for accuracy.
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Homework Statement



I am trying to solve the following homework problem:

1) A laser beam of light with wavelength λ=600nm shines perpendicular to a slide that has a
rectangular pattern of small holes that can be viewed as point sources. The spacing
between the holes is a=40μm and b=30μm along the x and y directions, respectively. The
light passing through the slide is projected on a screen 2 meters behind the slide.
Determine the positions of the spots on the screen using the small angle approximation.
Describe how the answer to problem 1 changes if the holes are squares of side 10μm.



Homework Equations



Sigma[e^i(Delta(k)*T]*S(g)

S(g) = integral[(e^(G dot r)*A(r))dr]

The Attempt at a Solution



For the first part, I solved for G, which is the basis for the reciprocal lattice. Since a1 = 40 um and a2 = 30 um, I used the formula which states that a(i) dot b(j) = 2Pi (kronecker delta i,j), so b1 = 2Pi / 40 and b2 = 2Pi / 30. I am confused how I use these reciprocal lattice vectors. Also, do I solve for S(g) in the first part as well as the second part?

Thanks
 
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  • #2
for your question! It seems like you're on the right track with using the small angle approximation to solve this problem. To answer your first question, the reciprocal lattice vectors are used to determine the positions of the spots on the screen. The spots will correspond to the diffraction pattern produced by the rectangular pattern of holes on the slide. You can use the small angle approximation to calculate the positions of the spots on the screen.

To answer your second question, you will need to solve for S(g) in both parts of the problem. In the first part, you will use the reciprocal lattice vectors to calculate the diffraction pattern produced by the rectangular holes. In the second part, you will need to recalculate S(g) using the new parameters for the square holes. This will give you the diffraction pattern for the square holes, which will have different positions on the screen compared to the rectangular holes.

I hope this helps clarify things for you. Keep up the good work!
 

What is 2-d Diffraction?

2-d Diffraction is a phenomenon in which waves, such as light or sound, spread out and bend around obstacles or through small openings. It is a fundamental property of waves and is an important concept in fields such as optics and acoustics.

How is 2-d Diffraction different from 1-d Diffraction?

1-d Diffraction occurs when a wave passes through a single slit or around a single obstacle, causing the wave to spread out in one direction. 2-d Diffraction, on the other hand, occurs when a wave passes through a slit or around an obstacle that is not only narrow in one direction, but also in the other direction, causing the wave to spread out in two dimensions.

What is the difference between 2-d Diffraction and 2-d Interference?

2-d Diffraction and 2-d Interference are often confused, but they are two distinct phenomena. 2-d Diffraction occurs when a wave passes through an opening or around an obstacle, while 2-d Interference occurs when two or more waves interact with each other. Both can result in the bending and spreading of waves, but they have different causes.

What factors affect the intensity of 2-d Diffraction patterns?

The intensity of 2-d Diffraction patterns is affected by several factors, including the wavelength of the wave, the size of the opening or obstacle, and the distance between the source of the wave and the obstacle. Additionally, the shape and material of the obstacle can also impact the intensity of the diffraction pattern.

How is 2-d Diffraction used in practical applications?

2-d Diffraction has many practical applications, particularly in the field of optics. It is used in the design of optical instruments such as microscopes and telescopes, as well as in the production of holograms. It is also used in acoustics to study sound waves and in the field of X-ray crystallography to determine the structure of molecules.

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