How Do You Derive the Diffraction Formula nλ = d sinθ?

In summary, the conversation discusses how to derive the formula n lambda=d sin theta and how to approach it when it hasn't been taught yet. It is suggested to use the similarity of triangles to combine the formulas and get the desired result.
  • #1
pianogirl
8
0

Homework Statement


Basically there's a question that I'm doing which asks me to show how to derive:
n lambda=d sin theta
We haven't learned it yet and so how do you do it?


Homework Equations


n lambda=d sin theta



The Attempt at a Solution


Not a clue!

Thanks!
 
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  • #2
  • #3


Hello,

Deriving the diffraction formula involves using principles from wave optics and trigonometry. First, we need to understand that when light passes through a narrow slit, it diffracts or bends around the edges of the slit. This creates a pattern of bright and dark spots, known as the diffraction pattern.

To derive the formula, we start by considering a single slit of width d. When light passes through this slit, it creates a wavefront that spreads out in all directions. However, due to the narrow width of the slit, the wavefronts become curved. This can be represented by a series of circular arcs, with the center of the slit as the center of each arc.

Next, we consider a point P on the screen where the diffraction pattern is formed. This point is a distance L away from the slit. We can draw a line from the top edge of the slit to point P, creating a right triangle. The angle made by this line with the horizontal is theta.

Using trigonometry, we can see that the length of the arc from the center of the slit to point P is equal to d sin theta. This is because the sine of an angle is equal to the opposite side (in this case, the distance from the center of the slit to point P) divided by the hypotenuse (the distance from the slit to point P).

Now, we know that the distance between bright spots in the diffraction pattern is equal to the wavelength of light, denoted by lambda. This means that the distance between the center of the slit and the first bright spot is also equal to lambda. Using this information, we can set up the following equation:

n lambda = d sin theta

Where n is the number of bright spots away from the center of the slit. This is known as the diffraction formula, and it tells us the relationship between the distance between bright spots, the wavelength of light, and the angle of diffraction.

I hope this explanation helps you understand how to derive the diffraction formula. It is important to have a good understanding of wave optics and trigonometry in order to fully understand the derivation. Let me know if you have any further questions.

Best,
 

Related to How Do You Derive the Diffraction Formula nλ = d sinθ?

1. What is diffraction and why is it important?

Diffraction is the bending of waves around obstacles or through small openings. It is important because it allows us to analyze and understand the behavior of waves in different situations, such as light passing through a narrow slit or sound waves around a corner.

2. How is the diffraction formula derived?

The diffraction formula is derived using Huygens' principle, which states that every point on a wavefront acts as a source of secondary wavelets. The formula takes into account the wavelength of the wave, the size of the opening or obstacle, and the distance between the source and the observation point.

3. What are the assumptions made in deriving the diffraction formula?

The diffraction formula assumes that the wavefronts are parallel and that the obstacle or opening is much smaller than the wavelength of the wave. It also assumes that the medium through which the wave is passing is homogeneous and isotropic.

4. How is the diffraction formula used in practical applications?

The diffraction formula is used in various fields, such as optics, acoustics, and radio waves. It is used to understand the behavior of waves in different scenarios, such as the diffraction of light in a microscope or the diffraction of sound in concert halls. It is also used in the design of devices like antennas and diffraction gratings.

5. Are there any limitations to the diffraction formula?

The diffraction formula has limitations in certain situations, such as when the size of the obstacle or opening is comparable to the wavelength of the wave. It also does not take into account factors like interference and polarization, which may affect the diffraction pattern. In these cases, more complex equations and models may be needed to accurately describe the diffraction phenomenon.

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