A) \[\omega \]
B) \[9\,\omega \]
C) \[4\,\omega \]
D) \[2\,\omega \]
Correct Answer: C
Solution :
In Young's double slit experiment interference width is given by fringe width \[(w)=\frac{\lambda D}{d}\] Where, d = sepration between the slits D = sepration between screen and slits Let \[w=\frac{\lambda D}{d}\] (Given) .....(i) In another case, d is taken as d/2 and D as 2D. \[\therefore \] \[w=\frac{\lambda \times (2D)}{\left( \frac{d}{2} \right)}=4\times \frac{\lambda D}{d}=4\omega \]You need to login to perform this action.
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