Q. A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ______ × 10⁻⁵ m.

Correct Answer: 429

Solution

In Young’s double slit experiment, the position of the nth bright fringe is given by:

y = nλD / d

For coincidence of bright fringes due to two wavelengths:

n₁λ₁ = n₂λ₂

Substituting λ₁ = 650 nm and λ₂ = 550 nm:

n₁ / n₂ = 550 / 650 = 11 / 13

The smallest integers satisfying this ratio are:

n₁ = 11, n₂ = 13

Least distance from central maximum is:

y = nλD / d

Using λ = 650 nm:

y = (11 × 650 × 10⁻⁹ × 1.2) / (2 × 10⁻³)

y = 4.29 × 10⁻³ m = 429 × 10⁻⁵ m

Hence, the required answer is 429.

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