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  • A red LED emits light at 0.1 watts uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode isOption: 1

A red LED emits light at 0.1 watts uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is

Option: 1

5.48 V/m


Option: 2

7.75 V/m


Option: 3

1.73 V/m


Option: 4

2.45 V/m


Answers (1)

best_answer

The intensity of light, \mathrm{\mathrm{I}=\mathrm{u}_{\mathrm{av}} \mathrm{c}}

Also,       \mathrm{\mathrm{I}=\mathrm{P} / 4 \pi \mathrm{r}^2 \text { and } \mathrm{u}_{\mathrm{av}}=1 / 2 \varepsilon_0 \mathrm{E}_0^2}

                     2                          2

\mathrm{\therefore \mathrm{P} / 4 \pi \mathrm{r}=(1 / 2) \varepsilon_0 \mathrm{E}_0 \mathrm{c}}

Or \mathrm{E_0=\sqrt{\frac{2 P}{4 \pi \epsilon_0 r^2 c}}}

Here, \mathrm{\mathrm{P}=0.1 \mathrm{~W}, \mathrm{r}=1 \mathrm{~m}, \mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}

\begin{aligned} & 1 / 4 \pi \varepsilon_0=9 \times 10^9 \mathrm{~N} \mathrm{C}^{-2} \mathrm{~m}^2 \\ \\& \quad \text{E}_0=\sqrt{\frac{2 \times 0.1 \times 9 \times 10^9}{1^2 \times 3 \times 10^8}}=\sqrt{6} \\ \\& =2.45 \mathrm{~V} \mathrm{~m}^{-1} \end{aligned}

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SANGALDEEP SINGH

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