A) \[\frac{{{l}_{0}}}{2}\]
B) \[{{l}_{0}}\]
C) \[2{{l}_{0}}\]
D) \[4{{l}_{0}}\]
Correct Answer: B
Solution :
Idea Here, students must remember that the resultant intensity in YDSE depends on the phase difference between the two waves Fringe width is\[\beta =\frac{\lambda D}{d}\] Given distance from central maxima\[=2.25\times \frac{\lambda D}{d}\] Also, we know that path difference\[\Delta x\]between two interfering waves at a distance y from the centre is given by \[\Delta x=\frac{yd}{D},\left[ \text{given y = 2}\text{.25}\frac{\lambda D}{d} \right]\] Hence, \[\Delta x=\frac{2.25\times D\lambda }{d}\times \frac{d}{D}\] \[=2.25\lambda \] Hence phase difference is\[\phi =\frac{2\pi }{\lambda }\times \Delta x\] \[=\frac{2\pi }{\lambda }\times 2.25\lambda \] \[=4.5\lambda \] Now, \[l={{l}_{\max }}{{\cos }^{2}}\phi /2\] \[={{l}_{\max }}\times {{\cos }^{2}}\left( \frac{4.5\pi }{2} \right)\] \[={{l}_{\max }}{{\cos }^{2}}\frac{\pi }{4}\] \[\Rightarrow l=\frac{{{l}_{\max }}}{4}\] Now, \[{{l}_{\max }}=4{{l}_{0}}\] So, \[l=\frac{4{{l}_{0}}}{4}={{l}_{0}}\] TEST Edge In YDSE, different types of questions could be asked. The questions based on dark and bright fringes could be asked.
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