in young's double slit experiment wavelength = 500nm, distance between the slits,d=1.0 mm, Dist of screen from the slits,D=1.0m. Find the minimum distance from the central maxima for which intensity is half of the maximum intensity

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Vikas TU · Answer

As we all know that From the given question Wavelength(w) = 500 nm D = 1 m, d = 10^-3 I = Imax/2 Y = 5 x 10^-7/10^-3 Y = 5x 10^-4

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in young's double slit experiment wavelength = 500nm, distance between the slits,d=1.0 mm, Dist of screen from the slits,D=1.0m. Find the minimum distance from the central maxima for which intensity is half of the maximum intensity

in young's double slit experiment wavelength = 500nm, distance between the slits,d=1.0 mm, Dist of screen from the slits,D=1.0m. Find the minimum distance from the central maxima for which intensity is half of the maximum intensity

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in young's double slit experiment wavelength = 500nm, distance between the slits,d=1.0 mm, Dist of screen from the slits,D=1.0m. Find the minimum distance from the central maxima for which intensity is half of the maximum intensity

in young's double slit experiment wavelength = 500nm, distance between the slits,d=1.0 mm, Dist of screen from the slits,D=1.0m. Find the minimum distance from the central maxima for which intensity is half of the maximum intensity

1 Answers

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