A) \[a=\sqrt{\lambda L}\] and \[{{b}_{\min }}\,=\sqrt{4\lambda L}\]
B) \[a=\frac{{{\lambda }^{2}}}{L}\] and \[{{b}_{\min \,}}=\sqrt{4\lambda L}\]
C) \[a=\frac{{{\lambda }^{2}}}{L}\] and \[{{b}_{\min }}=\left( \frac{2{{\lambda }^{2}}}{L} \right)\]
D) \[a=\sqrt{\lambda 1}\] and \[{{b}_{\min }}=\left( \frac{2{{\lambda }^{2}}}{L} \right)\]
Correct Answer: A
Solution :
[a] Given geometrical spread=a Diffraction spread \[=\frac{\lambda }{a}\times L=\frac{\lambda L}{a}\] The sum \[b=a+\frac{\lambda L}{a}\] For b to be minimum \[\frac{db}{da}=0\Rightarrow \frac{d}{da}\left( a+\frac{\lambda L}{a} \right)=0\Rightarrow a=\sqrt{\lambda L}\] \[{{b}_{\min }}=\sqrt{\lambda L}+\sqrt{\lambda L}=2\sqrt{\lambda L}=\sqrt{4\lambda L}\]You need to login to perform this action.
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