Modern Physics Ques 280
- A red LED emits light at $0.1$ $ W$ uniformly around it. The amplitude of the electric field of the light at a distance of $1 $ $m$ from the diode is
(2015 Main)
(a) $2.45 $ $ V / m$
(b) $1.73 $ $ V / m$
(c) $5.48 $ $ V / m$
(d) $7.75 $ $ V / m$
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Answer:
Correct Answer: 280.(a)
Solution:
- Intensity at a distance $r$ from a point source of power $P$ is given by
$ \begin{aligned} I & =\frac{P}{4 \pi r^{2}} \quad …….(i) \\ \text { Also, } \quad I & =\frac{1}{2} \varepsilon _0 E _0^{2} c \quad …….(ii) \end{aligned} $
where, $E _0$ is amplitude of electric field and $c$ the speed of light. Eqs. (i) and (ii) we get
$ \begin{aligned} E _0 & =\sqrt{\frac{2 P}{4 \pi \varepsilon _0 r^{2} c}} \\ & =\sqrt{\frac{2 \times 9 \times 10^{9} \times 0.1}{(1)^{2} \times 3 \times 10^{8}}} \\ & =\sqrt{6}=2.45 V / m \end{aligned} $
BETA
