Question
In Young's double slit interference experiment, the wavelength of light used is 6000˙A. If the path difference between waves reaching a point P on the screen is 1.5 microns, then at that point P
Answer: Option C
:
C
λ=6000˙A,x=1.5μm=15000˙A
For min. (dark band)
x=(2n−1)λ2;15000=(2n−1)60002
15000=(2n−1)3000
5+1=2n
n=62n=3
3rd dark band
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C
λ=6000˙A,x=1.5μm=15000˙A
For min. (dark band)
x=(2n−1)λ2;15000=(2n−1)60002
15000=(2n−1)3000
5+1=2n
n=62n=3
3rd dark band
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