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← JEE Main 2022 · 29 Jun · Shift 2
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JEE Main 2022 · 29 Jun · Shift 2, Physics question 59

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5×10−25\times10^{-2} m towards the slits, the change in fringe width is 3×10−33\times10^{-3} cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ______ nm.

Answer and worked solution

Official answer: 600

Change in fringe width. Δβ=λΔDd\Delta\beta=\dfrac{\lambda\Delta D}d: λ=3×10−5×10−35×10−2=6×10−7\lambda=\dfrac{3\times10^{-5}\times10^{-3}}{5\times10^{-2}}=6\times10^{-7} m =600=600 nm.

Answer: 600

Topics:Interference and Young's double slit
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