In Young's double slit experiment, the intensity on the screen at a point where path difference $ \lambda $ is $ \mathrm{K} $. What will be the intensity at the point where path difference is $ \frac{\lambda}{4} $ :

Question

In Young's double slit experiment, the intensity on the screen at a point where path difference $ \lambda $ is $ \mathrm{K} $. What will be the intensity at the point where path difference is $ \frac{\lambda}{4} $ :, $ \frac{K}{4} $, $ \frac{K}{2} $, $ \mathrm{K} $, Zero

MARKS App · Accepted Answer

By using phase difference ϕ = 2 π λ Δ For path difference λ , Phase difference ϕ 1 = 2 π and for path difference λ 4 , phase difference ϕ 2 = π 2 . Also by using I = 4 I 0 cos 2 ⁡ ϕ 2 ⇒ I 1 I 2 = cos 2 ⁡ ϕ 1 2 cos 2 ⁡ ϕ 2 2 ⇒ K I 2 = cos 2 ⁡ 2 π 2 cos 2 ⁡ π 2 = 1 1 2 ⇒ I 2 = K 2

Search any question & find its solution

Looking for more such questions to practice?