# Single-slit diffraction equations

1. May 10, 2007

### AznBoi

Please tell me what I need to know and what equations I need to use for these type of problems. Please define the variables in the equations too. Thanks.

-Sketch or identify the intensity patteren that results when monochromatic waves pass through a single slit and fall on a distant screen, and describe how this pattern will change if the slit width or the wavelength of the waves is changed and Calculate for a single-slit pattern, the angles or the positions on a distance screen where the intensity is zero.

What is intensity? Is it where the maximas (bright frindges) are? Is there zero intensity where there is a minima (dark fridge)?

I don't know if this is the right equation to use for the objective above:
$$sin \theta_{dark}=\frac{m\lambda}{a}$$

2. May 10, 2007

### AznBoi

I guess the reason why I'm confused because it is not like diffraction/interference with a double-slit since the equation used in a double-slit actually gives you the position of the bright and dark frindges (places where the waves interefere constructively and destructively) while the single slit equation can only give you the angle from the central maxima where waves interfere destructively.

Am I right about this? Is there an equation where you can find the position of the dark/bright frindges from the central maxima? Thanks

3. May 10, 2007

### maverick280857

You need to know how the intensity varies over the screen, in a single slit diffraction. For a single slit, the formula is:

$$I = I_{0}\frac{\sin^{2}\beta}{\beta^2}$$

where

$$\beta = \frac{\pi b}{\lambda}\sin\theta$$

where $b$ is the slit width, $\lambda$ is the wavelength of light used and $I_{0}$ is the intensity at the central maximum ($\theta = 0$), i.e. the maximum intensity. Without a proper figure, it is not possible to make this clearer.

You should refer to a physics textbook (Halliday/Resnick, Sears/Zemansky, etc.) at this stage. You might also want to have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

Hope this helps.

4. May 10, 2007

### maverick280857

Sorry, I didn't read your second post while I wrote my first.

If you denote by $D$ the distance from the source to the screen and $y$ the distance of any point from the central bright fringe on the screen, then

$$\sin\theta = \frac{y}{\sqrt{y^2+D^2}}$$

If y is small compared to D (as is generally the case for interference), you can write approximately

$$\sin\theta = \tan\theta = \frac{y}{D}$$

and you can plug this in for $\sin\theta$ and get $y$.

5. May 10, 2007

### AznBoi

So basically, the places where there is zero intensity have a dark frindge (destructive interference) and the places where their is maximum intensity have a bright frindge (constructive interference)?

6. May 10, 2007

### maverick280857

If you treat the slit as a source of several light 'beams'..(do you see what I am trying to get at, here?)

Did you see the hyperphysics link?

BTW, what textbook are you using?

[Sidenote: Studying interference before diffraction is natural, but when you switch to diffraction, you tend to think about it in the same way as interference. Thats how you can't explain the intensity pattern: in interference, the intensity doesn't fall off, so you tend to expect the same thing in diffraction. Strictly speaking, you shouldn't be neglecting diffraction effects in studying interference patterns...but when you do you get the simplified model that you already know so much about. Don't treat diffraction and interference as two totally distinct phenomena.]

7. May 10, 2007

### AznBoi

I'm using college physics 7th ed. by serway/faughn. The books says that "the values of theta for which the diffraction pattern has zero intensity, where a dark fringe forms. The various dark fringes (points of zero intensity) occur at the values of theta that satisfy $$Sin\theta_{dark}=\frac{m\lambda}{a}$$ a, being the width of the slit. So basically dark frindge=zero intensity, and bright frindge=max intensity?

8. May 11, 2007

### maverick280857

Yup, and you can see that from the expression for intensity I gave a few posts ago.

9. May 11, 2007

### AznBoi

Thanks a lot maverick. I really appreciate your help!

10. Apr 21, 2011

### Fjolvar

This is an old post, but just to clarify in case someone else has the same question in the future..

The bright fringes do not represent maximum intensity. Maximum intensity only occurs when theta is zero which is the bright fringe in the middle of the diffraction pattern. As theta is increased and you move further from the center of the diffraction pattern, the intensity decreases for each bright fringe. Just wanted to clarify that.