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JEE Main 2026 (28 Jan, Shift 2) · YDSE: Film, Medium & Two Wavelengths

YDSE: Film, Medium & Two Wavelengths question from JEE Main 2026 (28 Jan, Shift 2)

Q18JEE Main 2026 · 28 JAN · SHIFT 2Medium

YDSE: Film, Medium & Two Wavelengths

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\times 10^{-5} m.

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

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Answer: 429

  1. Coincidence: n1×650=n2×550⇒13n1=11n2n_1 \times 650 = n_2 \times 550 \Rightarrow 13n_1 = 11n_2, so n1=11n_1 = 11.
  2. y=11×650×10−9×1.22×10−3=4.29×10−3 my = \dfrac{11 \times 650\times10^{-9} \times 1.2}{2\times10^{-3}} = 4.29\times10^{-3}\text{ m}
  3. So the answer is 429×10−5429 \times 10^{-5} m, i.e. 429.
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