A) \[0.62\mu m\]
B) \[1.6\mu m\]
C) \[0.96\mu m\]
D) \[4.2\mu m\]
Correct Answer: C
Solution :
[c] : Intensity \[=\frac{1}{2}{{A}^{2}}{{\omega }^{2}}\rho v=\frac{1}{2}{{A}^{2}}{{(2\pi \upsilon )}^{2}}\rho v\]\[=2{{\pi }^{2}}{{\upsilon }^{2}}{{A}^{2}}\rho v\] Also, intensity\[=\frac{power}{4\pi {{r}^{2}}}\] \[\therefore \]\[2{{\pi }^{2}}{{\upsilon }^{2}}{{A}^{2}}\rho v=\frac{power}{4\pi {{r}^{2}}}\] \[A=\frac{1}{2\pi rv}\sqrt{\frac{power}{2\pi \rho v}}\] \[A=\frac{1}{2\pi (10)(1000)}\sqrt{\frac{10}{2\pi (1.29)340}}\] \[=9.6\times {{10}^{-7}}m=0.96\mu m\]
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