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In a modified set up of Young's double slit experiment, it is given that S S_(2)-S S_(1)=(lamda)/(4) i.e., the source S is not equidistant from the slits S_(1) and S_(2). (a) Obtain the condition for constructive and destructive interference at any point P on the screen in terms of the path difference Delta=S_(2)P-S_(1)P. (b) Does the observed central bright fringe lie above or below O ? Give reason in support of your answer. |
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Answer» Solution :(a) The Net path difference between two waves, reaching on the screen through two slits, is given as: `S S_(2)P-S S_(1)P=(S S_(2)-S S_(1))+(S_(2)P-S_(1) P)` It is given that `S S_(2)-S S_(1)=(lamda)/(4)` and we KNOW that `(S_(2)P-S_(1)P)=DELTA=(xd)/(D)`, where .d. is the separation between slits `S_(1) and S_(2) and .D.` the DISTANCE of screen from the double-slit. `therefore` Net path difference `=(lamda)/(4)+(xd)/(D)`  (i) For constructive interference `(lamda)/(4)+(xd)/(D)=nlamda`, where `N=0,1,2,3 . . .` `implies(xd)/(D)=nlamda-(lamda)/(4)=(4n-1)(lamda)/(4)implies(x_(n))_("bright")=((4n-1)lamdaD)/(4d)` (ii) For destructive interference `(lamda)/(4)+(xd)/(D)=(2n-1)(lamda)/(2)`, where n=1,2,3, . . `implies(xd)/(D)=(4n-3)(lamda)/(4)implies(x_(n))_(dark)=((4n-3)lamdaD)/(4d)`. (b) For central bright fringe n=0 and hence `(x_(0))_("bright")=-(lamdaD)/(4d)`. Thus, the observed central bright fringe shifts TOWARDS the line of slit `S_(2)`. |
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