In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 1/40 by using light of wavelength λ 1. When the light of wavelength λ 2 is used a bright fringe is seen at the same angle in the same set up. Given that λ 1 and λ 2 are in visible range (380 nm to 740 nm), their values are - (a) 400 nm, 500 nm (b) 625 nm, 500 nm (c) 380 nm, 500 nm (d) 380 nm, 525 nm

Answer: (b). Bright fringe: dsin θ ≈ dθ = nλ. dθ = (0.1×10⁻³)×1/40 = 2.5×10⁻⁶ m = 2500 nm λ = 2500/n nm: n = 4 ⇒ 625 nm, n = 5 ⇒ 500 nm, both in the visible range; only option (b) matches. So the answer is (b) 625 nm, 500 nm.

In a Young’s double slit experiment with slit separation 0.1 mm, one… | JEE Main 2019 (10 Jan, Shift 1)

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YDSE: Fringe Width & Fringe Positions

In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle $\frac{1}{40}$ by using light of wavelength $\lambda$ 1. When the light of wavelength $\lambda$ 2 is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda$ 1 and $\lambda$ 2 are in visible range (380 nm to 740 nm), their values are -

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Answer: (b)

Bright fringe: $d\sin\theta \approx d\theta = n\lambda$ .
$d\theta = (0.1\times10^{-3})\times\frac{1}{40} = 2.5\times10^{-6}\text{ m} = 2500\text{ nm}$
$\lambda = \frac{2500}{n}$ nm: $n = 4 \Rightarrow 625$ nm, $n = 5 \Rightarrow 500$ nm, both in the visible range; only option (b) matches.
So the answer is (b) 625 nm, 500 nm.

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