    Question

# In a YDSE apparatus, two identical slits are separated by 1 and distance between the slits and screen is 1m.The wavelength of light used is 6000A∘. The minimum distance between two points on screen having 75% intensity of the maximum intensity is :

A
0.9
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B
0.40
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C
0.30
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D
0.20
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Solution

## The correct option is A 0.9Given, Distance between slit and screen is D=1m Slit width is d=1m Wavelength of light is λ=6000A0=6×10−7m II0=75100=34Intensity of light in the fringe pattern is I=I0cos2(θ2)By equating 34I0=I0cos2(θ2)θ2=π12,5π6θ=π6,5π3Let △θ is path difference π6=πydλDy=λD6dSubstitute all value in above equation y=6×10−7×16×1=0.1μmAnd5π3=πydλDy=5λD3dSubstitute all value in above equation y=5×6×10−7×13×1=1μm△y=1−0.1=0.9μm  Suggest Corrections  0      Related Videos   Introduction to Separation
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