The distance of a point on the screen from two slits in biprism experiment is 1.8 × 10⁻⁵ m and 1.23 × 10⁻⁵ m. If wavelength of light used is 6000 Å the fringe number formed at that point is (a) 10 th bright (b) 10 th dark (c) 9 th bright (d) 9 th dark

Answer: (b). Path difference Δ = 1.8×10⁻⁵ - 1.23×10⁻⁵ = 5.7×10⁻⁶ m. (Δ)/(λ) = 5.7×10⁻⁶6×10⁻⁷ = 9.5, a half-integer, so the point is dark. Dark: Δ = (2n-1)(λ)/(2) ⇒ 2n-1 = 19 ⇒ n = 10. So the answer is (b) 10th dark.

The distance of a point on the screen from two slits in biprism… | MHT CET 2026 (14 May, Shift 2)

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Biprism, Thin Films & Other Devices

The distance of a point on the screen from two slits in biprism experiment is $1.8 \times 10^{-5}$ m and $1.23 \times 10^{-5}$ m. If wavelength of light used is $6000$ Å the fringe number formed at that point is

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Answer: (b)

Path difference $\Delta = 1.8\times10^{-5} - 1.23\times10^{-5} = 5.7\times10^{-6}\ \text{m}$ .
$\frac{\Delta}{\lambda} = \frac{5.7\times10^{-6}}{6\times10^{-7}} = 9.5$ , a half-integer, so the point is dark.
Dark: $\Delta = (2n-1)\frac{\lambda}{2} \Rightarrow 2n-1 = 19 \Rightarrow n = 10$ .
So the answer is (b) 10th dark.

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