path diffrenece at any point = 2*lamda*sin(theta)

for maxima path difference = n*lamda
hence. 2*lamda

*sin(theta)=n*lamda

find possible values of n

C-5 is answer

2 upar

2 neeche

1 central

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

Ek upar, ek niche, ek central

Arre bhai ne style copy ki

POTUS ki style kon copy nahi karna chahega

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.

Thank you all. Great explanations.

I was confused why answer was not 5. Understood.