A) 2
B) 1/2
C) 4
D) 16
Correct Answer: A
Solution :
\[I=4{{I}_{0}}{{\cos }^{2}}\frac{\varphi }{2}\] At central position \[{{I}_{1}}=4{{I}_{0}}\] ??(i) Since the phase difference between two successive fringes is \[2x,\] the phase difference between two points separated by a distance equal to one quarter of the distance between the two, successive fringes is equal to \[\delta =(2\pi )\,\left( \frac{1}{4} \right)=\frac{\pi }{2}\] radian \[\Rightarrow {{I}_{2}}=4{{I}_{0}}{{\cos }^{2}}\left( \frac{\frac{\pi }{2}}{2} \right)=2{{I}_{0}}\] ??(ii) Using (i) and (ii), \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{4\,{{I}_{0}}}{2\,{{I}_{0}}}=2\]
You need to login to perform this action.
You will be redirected in
3 sec
