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 \[6400-7900\,\overset{\text{o}}{\mathop{\text{A}}}\,,\] \[{{\lambda }_{V}}\](violet light) is \[4000-4500\,\overset{\text{o}}{\mathop{\text{A}}}\,\] \[\therefore \] \[{{W}_{R}}\approx 2{{W}_{V}}\]
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