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JEE Main 2026 · 4 Apr · Shift 2, Physics question 50

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of 7th bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is α×10−9\alpha\times10^{-9} m. The value of α\alpha is ________.

Answer and worked solution

Official answer: 5625

Shift of seven fringes. (μ−1)t=7λ(\mu-1)t=7\lambda, so
t=7×4500.56 nm=5625 nm=5625×10−9 m.t=\frac{7\times450}{0.56}\text{ nm}=5625\text{ nm}=5625\times10^{-9}\text{ m}.

Answer: 5625