A) \[\frac{n\pi }{2}+\frac{\pi }{4}+\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)\]
B) \[\frac{n\pi }{2}-\frac{\pi }{4}-\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)\]
C) \[\frac{n\pi }{2}+\frac{\pi }{4}+\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}-\frac{\theta }{2} \right)\]
D) \[\frac{n\pi }{2}+\frac{\pi }{4}+\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)\]
Correct Answer: A
Solution :
Let \[\Rightarrow \] \[M=(AL)d\,\,\,\Rightarrow \,\,\frac{M}{Ad}\] \[\Rightarrow \] \[T=2\pi \,\,\sqrt{\frac{M}{2Adg}}\] \[\Delta DBS,\] \[SD=\sqrt{{{60}^{2}}+{{25}^{2}}}\] \[=\sqrt{4225}=65=DP\] Now, \[\Delta x=(SA+AP)-SP\] \[\Rightarrow \] \[\Delta x=(65+65)-120\] \[\Rightarrow \] \[\Delta x=10\,m\] \[\frac{\lambda }{2}\] \[\lambda \] \[=\left( 10-\frac{\lambda }{2} \right)\] \[=(2n)\frac{\lambda }{2}\] \[n=0,\,\,1,\,\,2,...\] Also, \[10-\frac{\lambda }{2}=(2n)\frac{\lambda }{2},\,\,n=0,\,\,1,\,2,...\] \[10=(2n+1)\frac{\lambda }{2},\,n=0,\,1,\,\,2,...\] \[\Rightarrow \] \[\lambda =\frac{20}{2n+1},\,\,n=0,\,\,1,\,\,2,...\] \[\Rightarrow \] \[\lambda =20,\,\frac{20}{3},\,\frac{20}{5},\,\frac{20}{7},...\] \[\phi \] \[\tan \,\phi =\frac{V}{H}\] \[\theta \] \[H'=H\,\cos \,\theta ,\] (by using componendo and dividendo) \[\phi '\] \[\tan \,\phi =\frac{V}{{{H}^{.}}}=\frac{V}{H\,\cos \,\theta }\] \[\frac{\tan \,\phi '}{\tan \,\phi }=\frac{V/H\,\cos \,\theta }{\frac{V}{H}}=\frac{1}{\cos \,\theta }\] \[{{\phi }_{1}}=\] \[=BA\] \[{{\phi }_{2}}=\] \[||\] \[=0\] \[\therefore \,\,|\xi |\,=\,\left| \frac{-\Delta \phi }{\Delta t} \right|=\,\left| -\left( \frac{0-BA}{\frac{T}{4}-0} \right) \right|=\frac{4BA}{T}\] \[(\Delta \phi ={{\phi }_{2}}-{{\phi }_{1}})\]
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