JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Wave Optics

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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

Answer: 429

Solution

Bright fringes of the two wavelengths coincide when

n_{1} λ_{1} = n_{2} λ_{2}

. With

λ_{1} = 650

nm and

λ_{2} = 550

nm,

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

, so the least orders that coincide are

n_{1} = 11

and

n_{2} = 13

(since

11 \times 650 = 13 \times 550

). The position of this point is

y = \frac{n _{1} λ _{1} D}{d} = \frac{11 \times 650 \times 1 0 ^{- 9} \times 1.2}{2 \times 1 0 ^{- 3}} = 4290 \times 1 0^{- 6} m = 429 \times 1 0^{- 5} m

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Wave Optics

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