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# Path Difference Watch

1. http://filestore.aqa.org.uk/subjects...2-QP-JAN13.PDF

Question 7(e)

So my understand is that if you double the frequency, the wavelength will be halved right?

Also

Here is the MS
http://filestore.aqa.org.uk/subjects...W-MS-JAN13.PDF

I don't understand how they managed to get a path difference of 2 lambda? I know that if it is an integer then it will be constructive interference so therefore a maxima but how did they manage to get 2 here?

Also question 7B(II) is slightly confusing me. Why is the path difference equal to the wavelength?

Thanks
2. (Original post by CrazyFool229)
http://filestore.aqa.org.uk/subjects...2-QP-JAN13.PDF

Question 7(e)

So my understand is that if you double the frequency, the wavelength will be halved right?

Also

Here is the MS
http://filestore.aqa.org.uk/subjects...W-MS-JAN13.PDF

I don't understand how they managed to get a path difference of 2 lambda? I know that if it is an integer then it will be constructive interference so therefore a maxima but how did they manage to get 2 here?

Also question 7B(II) is slightly confusing me. Why is the path difference equal to the wavelength?

Thanks
The MS is a bit confusing... but the MS isn't a model answer. I guess it means 2x new lambda (i.e after the frequency is doubled) making it a maximum (but not the first maximum anymore)
3. (Original post by Joinedup)
The MS is a bit confusing... but the MS isn't a model answer. I guess it means 2x new lambda (i.e after the frequency is doubled) making it a maximum (but not the first maximum anymore)
But surely if the frequency has been doubled then the new wavelength will be 1/2 rather than 2? Unless I am being absolutely idiotic here which I probably am!
4. (Original post by CrazyFool229)
But surely if the frequency has been doubled then the new wavelength will be 1/2 rather than 2? Unless I am being absolutely idiotic here which I probably am!
Yeah - my understanding of the question is that the slits, position of the detector and everything else stays the same but that the source frequency is doubled.

that means the path difference that used to be 1x old lambda is now 2x new lambda (because the length of path difference stays the same but lambda has been halved)
5. (Original post by CrazyFool229)
But surely if the frequency has been doubled then the new wavelength will be 1/2 rather than 2? Unless I am being absolutely idiotic here which I probably am!
Yeah so the path difference will now be two lambda instead of one lambda which corresponds to constructive interference

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