In the youngs double slit experiment using a monochromatic light of wavelength , the path difference corresponding to any point having half the peak intensity is

Dear Student, 
According to Malus's Law
I=4I0Cos2(φ/2)and Acc to question, I=2I02I0=4I0Cos2(φ/2)Cos2(φ/2)=1/2Cos(φ/2) = 1/2φ/2=π/4φ=π/2

Thus, Path Difference x =λ2π×φx =λ2π×π2x =λ4

Regards

 

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