A) \[2\] times
B) \[3\] times
C) Equal
D) \[1/2\] times
Correct Answer: A
Solution :
The fringe width \[(W)\] is given by \[W=\frac{D\lambda }{d}\] where \[\lambda \] is wavelength, \[d\] the distance between coherent sources, \[D\] the distance between source and screen. \[\therefore \] \[\frac{{{W}_{R}}}{{{W}_{V}}}=\frac{{{\lambda }_{R}}}{{{\lambda }_{V}}}\] \[{{\lambda }_{R}}\](red light) is\[6400-7900{\AA}\] \[{{\lambda }_{V}}\](violet light) is\[4000-4500{\AA}\] \[\therefore \] \[{{W}_{R}}\approx 2{{W}_{V}}\]
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