A) zero
B) increased by \[6479\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) increased by \[589\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) decreased by \[589\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: C
Solution :
Angular fringe width \[\theta =\frac{\beta }{D}=\frac{\lambda }{2d}\] i.e., \[\theta \propto \lambda \] or \[\frac{\theta }{\theta }=\frac{\lambda }{\lambda }\] or \[\frac{\theta -\theta }{\theta }=\frac{\lambda -\lambda }{\lambda }\] Hence, percentage increase in wavelength = percentage increase in an angular fringe width =10% : or increase in wavelength \[=\frac{10}{100}\times 5890\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=589\overset{\text{o}}{\mathop{\text{A}}}\,\]
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