Question

In Young’s double slit experiment the maximum intensity of light is I_max, then the intensity of light of the path difference \(\frac{\lambda}{2}\) will be

Answer 1

(d) zero

At path difference \(\frac{\lambda}{2}\), destructive interference will occur. So intensity will be I_min
Let I₁ = I₂ = I₀
∴ I_max = I₁ + I₂ + \(2 \sqrt{I_{1} I_{2}}\) = I₀ + I_o + 2I₀ = 4I₀
and I_min = I₁+ I₂ \(2 \sqrt{I_{1} I_{2}}\) = I₀ + I_o – 2I₀ = 0