AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

According to figure, point P and point Q are at same phase.
According to figure,In △POR
cos‌θ=‌

PR

OP

=‌

d

OP

⇒‌‌OP=‌

d

cos‌θ

....(i)
In ∆QOP,
sin(90∘−2θ)=‌

OQ

OP

⇒‌‌cos‌2‌θ=‌

OQ

OP

⇒‌‌OQ=OP‌cos‌2‌θ....(ii)
∴ Path difference,
∆=OP+OQ=OP+OP‌cos‌2‌θ
=‌

d

cos‌θ

⋅2cos2θ  [from Eq. (i)]
=2d‌cos‌θ
But path difference is ‌

λ

2

.
∴‌‌2d‌cos‌θ=‌

λ

2

⇒‌‌cos‌θ=‌

λ

4d