JEE Main 2026 Jan 28 Shift 2, Physics Q24: Young’s Double Slit Experiment

JEE Main2026Jan 28, Shift 2Physics

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 ______ $\times 1 0^{- 5}$ m.

Your answer

Solution

For coincidence of bright fringes: $n_{1} λ_{1} = n_{2} λ_{2}$ .

$\frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{550}{650} = \frac{11}{13}$ .

Smallest integers: $n_{1} = 11, n_{2} = 13$ .

Position: $y = \frac{n _{1} λ _{1} D}{d} = \frac{11 \times 650 \times 1 0 ^{- 9} \times 1.2}{2 \times 1 0 ^{- 3}} = 4.29 \times 1 0^{- 3}$ m $= 429 \times 1 0^{- 5}$ m.

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JEE Main 2026 Jan 28 Shift 2, Physics Q24: Young’s Double Slit Experiment

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