Question

In Young's double slit experiment, d = 2 mm, D = 2 m, and λ = 500 nm. If intensity of two slits are I0 and 9I0, then find the intensity at y = 1/6 mm.

Answer 1

In Young's double slit experiment , path difference is given by
∆x = yd/D
Here y = 1/6 mm = 10⁻³/6 m
d = 2mm = 2 × 10⁻³ m
D = 2 m
∴ ∆x = 10⁻³/6 × 2 × 10⁻³/2 = 10⁻⁶/6 m

We know, the relation between phase difference and path difference
Ф = 2π/λ ∆x
= 2π/(5 × 10⁻⁷ ) × 10⁻⁶/6
= 2π/3 = 120°

Now, use resultant intensity formula,
e.g., I = $\bold{I_1+I_2+2\sqrt{I_1I_2}cos\Phi}$
Put Ф = 120°
I = I₀ + 9I₀ + 2√{I₀ × 9I₀} cos120°
= 10I₀ + 2√(3I₀)² × (-1/2)
= 10I₀ - 3I₀
= 7I₀

Hence, intensity at y = 1/6 mm

Answer 2

Answer:

refer the attachment

Explanation:

hope this helps you