A) 1.34 V/m
B) 2.68 V/m
C) 4.02 V/m
D) 5.36 V/m
Correct Answer: B
Solution :
| [b] Wavelength of monochromatic green light |
| \[=5.5\times {{10}^{-5}}cm\] |
| Intensity \[\text{I=}\frac{\text{Power}}{\text{Area}}\] |
| \[\text{=}\frac{100\times \left( 3/100 \right)}{4\pi {{\left( 5 \right)}^{2}}}\] \[\text{=}\frac{3}{100\pi }W{{m}^{-2}}\] |
| Now, half of this intensity (I) belongs to electric field and half of that to magnetic field, therefore, |
| \[\frac{I}{2}=\frac{1}{4}{{\varepsilon }_{0}}E_{0}^{2}C\] |
| or \[{{E}_{0}}=\sqrt{\frac{2I}{{{\varepsilon }_{0}}C}}\] |
| \[=\sqrt{\frac{2\times \left( \frac{3}{100}\pi \right)}{\left( \frac{1}{4\pi \times 9\times {{10}^{9}}} \right)\times \left( 3\times {{10}^{8}} \right)}}\] |
| \[=\sqrt{\frac{6}{25}\times 30}\] \[=\sqrt{7.2}\] |
| \[\therefore \]\[{{E}_{0}}=2.68V/m\] |
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