Manipal Entrance Test 2015 Physics Paper

Path difference at P is
Δx=2(

x

2

‌cos‌θ)=x‌cos‌θFor intensity to be maximum
Δx=nλ (n=0,1,2...)
⇒x‌cos‌θ=nλ
cos‌θ=

nλ

x

or cos‌θ≯1
∴

nλ

x

≯1
∴n≯

x

λ

Substituting x=5λ
n≯5 or n=1,2,3,4....
Therefore, in all four quadrants, there can be 20 maximas. There are more maximas at θ = 0° and θ = 180°.
But n = 5 corresponds to θ = 90° and θ = 270°, which are coming only twice. While we have multiplied it four times. Therefore, total numberof maximas still 20