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Central Diffraction Maximum Double Slit 
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#1
Apr1211, 03:59 PM

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1. The problem statement, all variables and given/known data
When a 450nm light is incident normally on a certain doubleslit system, the number of interference maxima within the central diffraction maxima is 5. When 900nm light is incident on the same slit system, the number of interference maxima within the central diffraction maxima is ______? 2. Relevant equations 3. 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? 


#2
Apr1211, 10:16 PM

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What about the width of the central diffraction maximum?
ehild 


#3
Apr1311, 09:46 PM

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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? 


#4
Apr1411, 12:13 AM

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Central Diffraction Maximum Double Slit
What is the central diffraction maximum? How its width depends on the wavelength?
Here is a link to study http://www.ualberta.ca/~pogosyan/tea...lecture36.html ehild 


#5
Apr1411, 09:19 AM

P: 1,089

"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? 


#6
Apr1411, 11:05 AM

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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 learnt something about diffraction by one slit and diffraction by multiple slits. ehild 


#8
Apr1411, 02:17 PM

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Yes.
ehild 


#9
Apr1411, 02:53 PM

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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
Apr1411, 03:15 PM

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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 


#12
Apr1511, 10:26 AM

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I am sorry, I can not solve the problem instead of you. I gave enough hints. The crucial point is the number of the interference maxima within the central diffraction maximum. You have to know something about diffraction by slits. Maybe, this helps. http://www.ualberta.ca/~pogosyan/tea...lecture35.html
ehild 


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