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In Young's double-slit experiment using monochromatic light of wavelength lamda, the intensity of light at a point on the screen where path difference is lamda, is K units. What is the intensity of light at a point where path difference is lamda//3 ? |
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Answer» Solution :In Young.s double-slit EXPERIMENT two SLITS are identical and intensity of light coming out of them is exactly same i.e., `I_(1)=I_(2)=I` (say). At a point on the screen where path difference is `lamda`, the phase difference `phi=(2pi)/(lamda),lamda=2pi" rad"` and HENCE resultant intensity of light there will be `I_(res)=4I"COS"^(2)(phi)/(2)=4Icos^(2)pi=4I=K` At the other point, where path difference is `(lamda)/(3)`, the corresponding phase difference `phi.=(2pi)/(lamda)xx(lamda)/(3)=(2pi)/(3)`, rad and hence resultant intensity at this point will be `I_(red).=4Icos^(2)((2pi)/(3))=4I(-(1)/(2))^(2)=I` Comparing (i) and (ii), we GET `(I_(res).)/(I_(red))=(I_(res).)/(K)=(I)/(4I)=(1)/(4)impliesI_(red).=(K)/(4)`. |
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