Central Diffraction Maximum Double Slit

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SUMMARY

The discussion revolves around the calculation of interference maxima within the central diffraction maximum for a double-slit system when illuminated by light of different wavelengths. When 450-nm light is used, there are 5 interference maxima. For 900-nm light, the number of maxima is derived from the relationship between the wavelengths and the slit parameters, leading to the conclusion that the number of maxima is 3. The key equations referenced include dsinθ = mλ and dsinθ = nλ', which relate the slit separation and wavelength to the angles of maxima.

PREREQUISITES
  • Understanding of double-slit interference patterns
  • Knowledge of diffraction principles, specifically central diffraction maximum
  • Familiarity with the equations dsinθ = mλ and dsinθ = nλ'
  • Basic grasp of how wavelength affects diffraction patterns
NEXT STEPS
  • Study the relationship between wavelength and the width of the central diffraction maximum
  • Learn about the effects of slit width and separation on interference patterns
  • Explore the concept of diffraction grating and its equations
  • Investigate the derivation of the number of maxima in double-slit experiments
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding interference and diffraction phenomena in wave mechanics.

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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.

dsin\theta = m \lambda

dsin\theta = n \lambda^{'}

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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What about the width of the central diffraction maximum?

ehild
 
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?
 
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?
 
Last edited by a moderator:
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
 
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 referring to?

See figure attached.
 

Attachments

  • Equation2slit.JPG
    Equation2slit.JPG
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Yes.

ehild
 
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?
 

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