Modern Physics 7 Question 32

33. 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) $7.75 V / m$

Show Answer

Solution:

  1. Intensity at a distance $r$ from a point source of power $P$ is given by

$$ \begin{aligned} I & =\frac{P}{4 \pi r^{2}} \\ \text { Also, } \quad I & =\frac{1}{2} \varepsilon _0 E _0^{2} c \end{aligned} $$

where, $E _0$ is amplitude of electric field and $c$ the speed of light. Eqs. (i) and (ii) we get

$$ \begin{aligned} E _0 & =\sqrt{\frac{2 P}{4 \pi \varepsilon _0 r^{2} c}} \\ & =\sqrt{\frac{2 \times 9 \times 10^{9} \times 0.1}{(1)^{2} \times 3 \times 10^{8}}} \\ & =\sqrt{6}=2.45 V / m \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane