# In a Young’s double slit experiment λ = 500 nm, d = 1.0 mm and D = 1.0 m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Jitender Pal
10 years ago
Sol. Given that, d = 2 mm = 2 * 10^–3 m, λ = 600 nm = 6 * 10^–7 m, I base max = 0.20 W/m^2, D = 2m For the point, y = 0.5 cm We know, path difference = x = yd/D = 0.5 * 10^-2 * 2 * 10^-3/2 = 5 * 10^-6 m So, the corresponding phase difference is, ∅ = 2πx/λ = 2π * 5 * 10^-6/6 * 10^-7 ⇒ 50π/3 = 16π + 2π/3 ⇒ ∅ = 2π/3 So, the amplitude of the resulting wave at the point y = 0.5 cm is, A = √r^2 + r^2 + 2r^2 cos(2π / 3) = √r^2 + r^2 – r^2 = r Since, l/l base max = A^2/(2r)^2 [since, maximum amplitude = 2r] ⇒ l/0.2 = A^2/4r^2 = r^2/4r^2 ⇒ l = 0.2/4 = 0.05 W/m^2.
4 years ago
Dear student,

Let the maximum intensity occuring at the central maximum be, Imax = Io
Let the minimum distance at which intensity, I = Io/2 be x
Now, I = Io cos2(ɸ/2)
hence, Io/2 = Io cos2(ɸ/2)
Hence, cos(ɸ/2) = 1/2
or, ɸ/2 = π/4
hence, π/λ * dx/D = π/4
Hence, x = λD/4d = 500 x 10-9 x 1/( 4 x 10-3)
or, x = 1.25 x 10-4 m

Thanks and regards,
Kushagra