Practice portal › Wave Optics › Interference and Young's Double Slit Experiment

A beam of light consisting of two wavelengths 7000 Å and 5500 Å is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is n×10⁻⁵ m. The value of n is ______.

Asked in JEE Main 6th April 2nd Shift 2023 · Fringe width and fringe position

Answer: 462

Step-by-step solution

Coincidence: n₁×7000 = n₂×5500 ⇒ 14n₁ = 11n₂, so n₁ = 11.

y = (11λ₁ D)/d = (11×7×10⁻⁷×1.5)/(2.5×10⁻³).

y = 4.62×10⁻³ m = 462×10⁻⁵ m.

→ 462

More Interference and Young's Double Slit Experiment questionsAll Interference and Young's Double Slit Experiment questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer