A red $LED$ emits light at $0.1 \mathrm{watt}$ uniformly around it. The amplitude of the electric field of the light at a distance of $1 \mathrm{m}$ from the diode is:

• A

  $7.75 \mathrm{V} {\mathrm{m}}^{-1}$
• B

  $1.73 \mathrm{V} {\mathrm{m}}^{-1}$
• Correct answer
  C

  $2.45 \mathrm{V} {\mathrm{m}}^{-1}$
• D

  $5.48 \mathrm{V} {\mathrm{m}}^{-1}$

Solution

For a point source of power = P, then intensity at a point at a separation x from the source is, $I=\frac{Power}{Area}=\frac{P}{4\pi x^2}$$\because$ The average intensity of E.M. wave is given by, $I=\frac{1}{2}c{\epsilon }_0{E}_o^2$$\Rightarrow E_0=\sqrt{\frac{2P}{4\pi {\epsilon }_{0}c x^2}}$

$\because \frac{1}{4\pi {\epsilon }_0}=9\times {10}^9, P=0.1 \mathrm{W}, x=1 \mathrm{m}$, $c=3\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}$

$\Rightarrow E_0=\sqrt{\frac{2\times 0.1\times 9\times {10}^9}{3\times {10}^8\times 1^2}}=\sqrt{6}=2.45 \mathrm{V} {\mathrm{m}}^{-1}$

Hence, the correct answer is option: 'C'.