 # Doubt from ydse

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path diffrenece at any point = 2lamdasin(theta)
for maxima path difference = nlamda
hence. 2
lamdasin(theta)=nlamda
find possible values of n

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2 upar
2 neeche
1 central

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Path difference at centre is 2lambda and at infinity is zero. B-3 would be answer.

Ek upar, ek niche, ek central

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Arre bhai ne style copy ki 1 Like   POTUS ki style kon copy nahi karna chahega

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Consider the intersection of the perpendicular lines to be O.

Now consider a point P at +infinity. What will be S1P-S2P? Almost 0. But it won’t be exactly 0. It will be a similar case when you consider a point Q at -infinity.

Now consider point O. What will be OS1-OS2? 2×lambda.

Now move from P to O to Q. At P path diffierence is almost 0. But not 0. So don’t consider it as a maxima. Then as you go down path diffierence increases from 0(×) to lamda to 2×lamda.

When you go from O to Q, you encounter another point where path diffierence will be lamda. ( Again don’t consider Q as maxima)

Thus in total you get 3 positions of maxima.

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Thank you all. Great explanations.
I was confused why answer was not 5. Understood.

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