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Consider the arrangement shownin figure. By some mechanism,the separation between the slits S_3 and S_4 can be changed. The intensity in measured at the point P which is at the common perpendicular bisector of S_1S_2 and S_3S_4. When z=(Dlamda)/(2d) the intensity measured at P is I. Find this intensity when z is equal to a. (Dlamda)/d, b. (3Dlamda)/(2d), c. (2Dlamda)/d |
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Answer» <P> So, `OS_3=OS_4=(lamdaD)/d` `rarr Dark fringe at `S_3 and S_4` `rarr At S_3` intensity at `S_3=0` `rarrl_1=0` At `S_4 Intensity at S_4=0` `rarr `l_2=0` `rarr At P PATH DIFFERENCE =0 `rarr PHASE difference =0` `rarr l=l_1+l_2+2sqrt(l_1l_2)COSTHETA^@=0+0+0=0` `rarr intensilty at P=0` |
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