Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm. Numerical answer type. Work it out, then open the answer.

Answer: 48. Coincidence: n₁ × 800 = n₂ × 600 ⇒ n₁ = 3, n₂ = 4. y = 3 × 800×10⁻⁹ × 70.35×10⁻³ = 0.048 m So the distance is 48 mm.

Two light waves of wavelengths 800 and 600 nm are used in Young's… | JEE Main 2023 (31 Jan, Shift 2)

Q48JEE Main 2023 · 31 JAN · SHIFT 2Difficult

YDSE: Film, Medium & Two Wavelengths

Two light waves of wavelengths 800 and $600\text{ nm}$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7\text{ m}$ away from plane of slits. If the two slits are separated by $0.35\text{ mm}$ , then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ $\text{mm}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 48

Coincidence: $n_1 \times 800 = n_2 \times 600 \Rightarrow n_1 = 3,\ n_2 = 4$ .
$y = \dfrac{3 \times 800\times10^{-9} \times 7}{0.35\times10^{-3}} = 0.048\text{ m}$
So the distance is 48 mm.

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