A) \[\frac{{{\gamma }_{1}}-{{\gamma }_{2}}}{3}+\alpha \]
B) \[{{E}_{I}}>{{E}_{II}}>{{E}_{III}}>{{E}_{IV}}>{{E}_{V}}\]
C) \[{{E}_{I}}>{{E}_{III}}={{E}_{V}}\,and\,\,\,{{E}_{II}}>{{E}_{IV}}\]
D) \[{{E}_{II}}={{E}_{IV}}={{E}_{V}}\,and\,\,\,{{E}_{I}}>{{E}_{III}}\]
Correct Answer: D
Solution :
For 5th dark fringe, \[{{x}_{1}}=(2n-1)\frac{\lambda }{2}\frac{D}{d}\] \[=\frac{9\lambda D}{2d}\] For 7th bright fringe, \[{{x}_{2}}=n\lambda \frac{D}{d}\] \[=\frac{7\lambda D}{d}\] Now. \[{{x}_{2}}-{{x}_{1}}=(\mu -1)t\frac{D}{d}\] \[\frac{\lambda D}{d}\left[ 7-\frac{9}{2} \right]=(\mu -1)t\frac{D}{d}\] \[t=\frac{25\lambda }{(\mu -1)}\]You need to login to perform this action.
You will be redirected in
3 sec