Question

Two light waves of the same frequency and constant phase difference have intensities in the ratio 81:1. When these waves are superposed, they produce inerference fringes. Calculate the ratio of maximum to minimum intensity

Answer 1

I₁/I₂=(a₁/a₂)²=81/1

a₁/a₂=√(81/1)=9/1

a1=9× a2

I_max / I_min
= (a₁+a₂)² / (a₁-a₂)²
=(9a2+a2)^2 / (9a2-a2)^2
=100 a^2 / 64 a^2
= 100/64

So, I_max : I_min= 100:64

Like if satisfied