in young'sdouble slit experiment wavelength = 500nm, distance between the slits,d=1.0 mm, Dist of screen from the slits,D=1.0m. Find theminimum distance from the central maxima for which intensity is half of the maximum intensity
Nikkitaa , 7 Years ago
Grade 12
1 Answers
Vikas TU
Last Activity: 7 Years ago
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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