# Diffraction Grating question

1. Sep 15, 2007

### nirvana1990

I'm doing an A level past paper for physics and there's a diagram involved which I can't put on here but here is the link:

http://www.egsphysics.co.uk/files/web_pages/exams.html [Broken]
It is the June 2004 paper unit 4 question 1.

The question is asking for the path difference between two points (C and E) on the same ray and I assumed it would be 2 wavelengths because this ray gives the 2nd order image but the mark scheme says it is 4 wavelengths.

Could anybody tell me why the path difference is 4 wavelengths and not 2?

Last edited by a moderator: May 3, 2017
2. Sep 15, 2007

### Kurdt

Staff Emeritus
Its simply because the triangle is drawn from 2 slits away. So B to D is two wavelengths and then the line carries on to E which makes C to E four wavelengths. I hope thats clear.

3. Sep 15, 2007

### nirvana1990

The triangle's drawn two slits away from what? From the zero order?
Are they only showing us the 2nd, 3rd and 4th orders and would the path difference between the 0 and 1st orders be 1 wavelength, 0 to 2nd would be 2 wavelengts etc. or have I got that completely wrong?!?!?

4. Sep 15, 2007

### Kurdt

Staff Emeritus
The description above the diagram says that light wave diffracted through an angle $\theta$ form 2nd order image. That means that the path difference between the rays is 2 wavelengths. On the diagram they illustrate this with a dotted line that forms a right angled triangle one ray as the smallest side. From trigonometry we get the formula $2\lambda = d sin (\theta)$. But on the diagram they extend this line to the next ray at point E so you have added on another two wavelengths. if you were to draw the right triangle from the slit below and call the new point E' then C to E' would be two wavelengths.

My descriptions aren't particularly good I apologise about that. But when I said it was drawn from two slits away I meant from the slit labelled C. Anyway its all about how the formula $n\lambda = d sin (\theta)$ is derived.

5. Sep 15, 2007

### nirvana1990

Oh right I wasn't understanding all this 2nd order image stuff! So what you're saying is that since it says i forms a 2nd order image after going through that angle, the path difference between each ray shown is 2 wavelengths? Does this mean that if they said a 1st order image was produced after going through this angle the path difference between each ray would be 1 wavelength?

Thanks for the help!

6. Sep 15, 2007

### Kurdt

Staff Emeritus
Yes it does. So if that angle was for a 1st order image c to E would be 2 wavelengths and B to D would be 1 wavelength.

7. Sep 15, 2007

### nirvana1990

Thanks sooo much again!