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Grade: 10
        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.
5 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
							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.

						
5 years ago
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