# Homework Help: Diffraction Grating, Light with two wavelengths

1. Jan 21, 2010

### AlmonzoWilder

1. The problem statement, all variables and given/known data
Light with wavelengths of 520 nm and 630 nm passes through a diffraction grating containing 6000 lines/cm.

a. Sketch a diagram of the image produced from m=0 to m=2, labeling the order of each fringe.

b. Calculate the angles for the first and second order maxima that would appear on the screen.

2. Relevant equations
sin$$\vartheta$$ = m$$\lambda$$/d

3. The attempt at a solution
If 6000lines/cm
Then,
d=1m/(6000linesx100cm/m)
d=1.6x10^-6 m/line
I'm not quite sure where to start, I don't need someone to solve it for me...just to help point me in the right direction, please?

2. Jan 21, 2010

### kuruman

You start from the "relevant equation" you have provided. You know the wavelength(s), you know d. To find the angle (from the sine), you need m. What is m for the first order maximum, what is m for the second order maximum?

3. Jan 21, 2010

### AlmonzoWilder

Okay, I follow to a certain extent...which wavelength do I use? Do I average them?

4. Jan 21, 2010

### kuruman

5. Jan 22, 2010

### AlmonzoWilder

No, I know what diffraction is. In my book the explanation and all the examples they give they only use one wavelength. I just don't know what to do with the two...calculate them induvidually?

6. Jan 22, 2010

### kuruman

If you know what a diffraction grating does, then you should know what happens when you shine light that consists of two or more wavelengths. What does happen? Before you focus on a formula and what to put in it, it is a good idea to have a picture in your mind about what is going on here.

7. Jan 22, 2010

### RoyalCat

The principle of superposition was used extensively to develop the theory behind diffraction. Try and use it to understand what's happening when you shine light through that consists of several wavelengths.

8. Jan 22, 2010

### AlmonzoWilder

Okay, I get what you are saying now, the difraction grating is used to separate the different wavelengths.
So I use the equation for each wavelength because the different wavelengths mean that they will have different angles for m=1, m=2 etc.
Am I on the right track or completely wrong?

9. Jan 22, 2010

### RoyalCat

That's exactly correct. :) As can be seen in the equation, the angles of diffraction maxima are completely wavelength dependent.
Physically, this means that for a given path length, some wavelengths undergo constructive interference (When the path difference is an integer multiple of that particular wavelength) and some undergo destructive interference, and some interfere in a way that's in between.

10. Jan 25, 2010

### AlmonzoWilder

Okay, so wave interference will only appear when what?
Thanks so much for your help, I really appreciate it!