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Question
In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ=500nm is incident on the slits. The total number of bright fringes that are observed in the angular range −30o≤θ≤30o is :
In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ=500nm is incident on the slits. The total number of bright fringes that are observed in the angular range −30o≤θ≤30o is :
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Solution
The correct option is B 641
Pam difference
dsinθ=nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32×10−3sin30=n×500×10−9
n=320
Hence total no. of maximas observed in angular range −30o≤θ≤30o is
maximas = 320 + 1 + 320 = 641
Pam difference
dsinθ=nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32×10−3sin30=n×500×10−9
n=320
Hence total no. of maximas observed in angular range −30o≤θ≤30o is
maximas = 320 + 1 + 320 = 641
dsinθ=nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32×10−3sin30=n×500×10−9
n=320
Hence total no. of maximas observed in angular range −30o≤θ≤30o is
maximas = 320 + 1 + 320 = 641
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