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← JEE Main 2022 · 24 Jun · Shift 1
PhysicsInteger+4 / -1Question 55 of 90

JEE Main 2022 · 24 Jun · Shift 1, Physics question 55

Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is ______ ×10−5\times10^{-5} m.

Answer and worked solution

Official answer: 3

First secondary maximum. asin⁡θ≈3λ2a\sin\theta\approx\dfrac{3\lambda}2, so y=3λD2ay=\dfrac{3\lambda D}{2a}.

Separation. Δy=3D Δλ2a=3×2×5×10−92×5×10−4=3×10−5\Delta y=\dfrac{3D\,\Delta\lambda}{2a}=\dfrac{3\times2\times5\times10^{-9}}{2\times5\times10^{-4}}=3\times10^{-5} m.

Answer: 3