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# Waves - young's fringe experiment watch

1. Can someone explain how to do this ppq I understand the first one but not the others. The mark scheme didn't help either.
Thanks
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2. (Original post by Exotic-L)
Can someone explain how to do this ppq I understand the first one but not the others. The mark scheme didn't help either.
Thanks
(i) Path difference is the difference in distance between a point and each of the two slits. The distance between Q and the first slit, S1, is 350mm, as shown in the diagram. The difference of Q from S2 requires the use of Pythagoras as this is a right angled triangle and is therefore sqrt(1202 + 3502) = 370mm. Therefore the path difference, S2Q - S1Q = 370mm - 350mm = 20mm.

(ii) Remember that maxima occur when there is constructive interference, hence the path difference must be an integer multiple of wavelengths, ie nλ, and P is the equidistant from both slits.

(iii) Use equation λ = ax/D, where a is the distance between the two slits, x is the distance between the fringes and D is the perpendicular distance of the screen from the slits.

Hope that helps, feel free to ask any more questions if you need any more help.

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