    Question

# In Young's double slit interference experiment the wavelength of light used is 6000A. If the path difference between waves reaching a ponit P on the screen is 1.5 microns, then at that point P:

A
second bright band occurs
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B
second dark band occurs
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C
third dark band occurs
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D
thirdbright band occurs
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Solution

## The correct option is C third dark band occursGiven λ=6000A=6000×10−10m=6×10−7mand △=1.5 microns=1.5×10−6m. For bright fringes: △=nλ; where n is an integer. n=△λ=1.5×10−66×10−7=52,which is not an integer. hence, path difference of 1.5×10−6m does not correspond to a bright fringe. For dark fringes, we have △=(n−12)λor 1.5×10−6=(n−12)×(6×10−7) which gives n−12=52 or n=3., Hence a path difference of 1.5×10−6m corresponds to the third dark fringe.  Suggest Corrections  0      Similar questions  Related Videos   YDSE Problems
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