Central Diffraction Maximum Double Slit

In summary: See figure attached.The number of interference maxima within the central diffraction maximum changes with the wavelength if everything stays the same: the width of the slits (a) and their separation (d).
  • #1
jegues
1,097
3

Homework Statement



When a 450-nm light is incident normally on a certain double-slit system, the number of interference maxima within the central diffraction maxima is 5. When 900-nm light is incident on the same slit system, the number of interference maxima within the central diffraction maxima is ______?

Homework Equations





The Attempt at a Solution



The answer is given as 5, but I can't figure out why.

[tex]dsin\theta = m \lambda[/tex]

[tex]dsin\theta = n \lambda^{'}[/tex]

Since dsin(theta) shouldn't change,

n = (m/2)

If we have 5 bright fringes in our central envelope, m = 2, therefore n=1.

So the number of bright fringes in the new central envelope is 2n+1 = 3 which is incorrect.

What am I misunderstanding?
 
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  • #2
What about the width of the central diffraction maximum?

ehild
 
  • #3
ehild said:
What about the width of the central diffraction maximum?

ehild


I don't know? Can you explain what you mean to me?

I've shown you everything I could think of up to this point. Can you give me another nudge in the right direction?
 
  • #5
ehild said:
What is the central diffraction maximum? How its width depends on the wavelength?
Here is a link to study
http://www.uAlberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect36/lecture36.html

ehild

Here's what I could find off that site that might be relevant to what we're doing. It's still all pretty confusing though,

"the light of different wavelength after passing through diffraction grating will have peaks of intensity at different angles,

θm(λ) ≈ m λ/d

producing the image something like this one for atomic hydrogen emission"

The other note I could has to do with a diffraction grating, even though we aren't dealing with a diffraction grating problem, maybe I can somehow relate it to the wavelength,

"maxima become narrower with more slits in the grating, hence the width of the maxima,

Δ sinθ ≈ λ/(N d)"

I'm still really confused. Am I on the right track?
 
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  • #6
Scroll down the page and find the relevant equation for two slits.

As the problem is about the number of interference maxima within the central diffraction maximum, you must have learned something about diffraction by one slit and diffraction by multiple slits.

ehild
 
  • #7
ehild said:
Scroll down the page and find the relevant equation for two slits.

As the problem is about the number of interference maxima within the central diffraction maximum, you must have learned something about diffraction by one slit and diffraction by multiple slits.

ehild


Is this the equation you were reffering to?

See figure attached.
 

Attachments

  • Equation2slit.JPG
    Equation2slit.JPG
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  • #8
Yes.

ehild
 
  • #9
ehild said:
Yes.

ehild

Okay but this is all still very confusing. This equation has a bunch of variables and things that haven't been given in the question. What sort of conclusion am I supposed to draw from this?

This equation refers to the intensity, correct?

The only thing I could note is that the coefficients of the sin terms will be getting smaller due to the increase in lambda.
 
  • #10
You are supposed to draw the conclusion how the number of interference maxima within the central diffraction maximum changes with the wavelength if everything stays the same: the width of the slits (a) and their separation (d).

ehild
 
  • #11
ehild said:
You are supposed to draw the conclusion how the number of interference maxima within the central diffraction maximum changes with the wavelength if everything stays the same: the width of the slits (a) and their separation (d).

ehild

So how do I figure out how it has changed?
 
  • #12

1. What is the central diffraction maximum in a double slit experiment?

The central diffraction maximum in a double slit experiment refers to the bright spot at the center of the interference pattern that is created when light or other waves pass through two parallel slits. It is a result of constructive interference, where the waves from each slit overlap and reinforce each other.

2. How is the central diffraction maximum affected by the distance between the slits in a double slit experiment?

The distance between the slits in a double slit experiment affects the central diffraction maximum by changing the spacing between the bright spots in the interference pattern. As the distance between the slits decreases, the central maximum becomes wider and more intense. As the distance increases, the central maximum becomes narrower and less intense.

3. What role does the wavelength of the waves play in the central diffraction maximum of a double slit experiment?

The wavelength of the waves plays a crucial role in the central diffraction maximum of a double slit experiment. The spacing between the bright spots in the interference pattern is directly proportional to the wavelength of the waves. This means that as the wavelength increases, the spacing between the bright spots also increases, resulting in a wider central maximum.

4. Can the central diffraction maximum be observed with particles instead of waves in a double slit experiment?

No, the central diffraction maximum cannot be observed with particles in a double slit experiment. The interference pattern is a result of the diffraction of waves, and particles do not exhibit diffraction. Instead, they will form two distinct bands behind the slits, with no central maximum.

5. Is the central diffraction maximum the only bright spot in a double slit interference pattern?

No, the central diffraction maximum is not the only bright spot in a double slit interference pattern. There are also secondary maxima and minima, which occur due to the interference of waves from the two slits. The central maximum is the brightest and most prominent spot, but the other maxima and minima play important roles in the overall interference pattern.

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