alleryFor YDSE with λ = 500 nm, d = 0.2 mm, D = 1 m, fringe width is:
A2.5 mm
B1 mm
C2.5 mm (β = 500×10⁻⁹ × 1 / 0.2×10⁻³ = 2.5 × 10⁻³ m)
D5 mm (= 500e-9 × 1 / 0.2e-3 × 2 — wait)
Answer & Solution
Correct answer: C. 2.5 mm (β = 500×10⁻⁹ × 1 / 0.2×10⁻³ = 2.5 × 10⁻³ m)
We have β = λD/d.
That is β = (500 × 10⁻⁹ × 1)/(0.2 × 10⁻³).
That is β = 2500 × 10⁻⁶ m.
That is β = 2.5 × 10⁻³ m.
So β = 2.5 mm.
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