# Central Diffraction Maximum Double Slit

1. Apr 12, 2011

### jegues

1. The problem statement, all variables and given/known data

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 ______?

2. Relevant equations

3. 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?

2. Apr 12, 2011

### ehild

What about the width of the central diffraction maximum?

ehild

3. Apr 13, 2011

### jegues

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. Apr 14, 2011

### ehild

Last edited: Apr 14, 2011
5. Apr 14, 2011

### jegues

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?

6. Apr 14, 2011

### ehild

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

7. Apr 14, 2011

### jegues

Is this the equation you were reffering to?

See figure attached.

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• ###### Equation2slit.JPG
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8. Apr 14, 2011

### ehild

Yes.

ehild

9. Apr 14, 2011

### jegues

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. Apr 14, 2011

### ehild

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. Apr 15, 2011

### jegues

So how do I figure out how it has changed?

12. Apr 15, 2011