What is the Path Difference for a Diffraction Grating?

[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
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?

 

[Kurdt]

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 that's clear.

 

[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?

 

[Kurdt]

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.

 

[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!

 

[Kurdt]

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!

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.

 

[nirvana1990]

Thanks sooo much again!