Optics

BHAI YEH YAHAN KA HIT QUES HAI
IT WAS IN THE FULL TEST 1
AND ALSO IN 1 OF THE TESTS IN THE PHYSICS DEPARTMENT (par maine donno bar galat kar diya )[17][17][17]

THE SOLUTION FOLLOWS

path difference Δx=CO+OP+λ/2(due to reflection)

=OP sin(90-2θ)+OP+λ/2=dsecθcos2θ+d secθ+λ/2=d secθ(1+cos2θ)+λ/2
=2dcosθ+λ/2
for maxima Δx=nλ=2dcosθ+λ/2

2dcosθ=(2n-1)λ/2
cosθ=(2n-1)λ/4d=λ/4d (for n=1)

I don't understand why λ/2 has been added to the path difference,can you please explain it?

CONSTRUCTIVE INTERFERENCE IS MENTIONED Mr.Sparkle

because of the reflection at the surface of a denser medium brings an additional phase change of Πor path diff of λ/2