A red LED emits light at 0-1 watt uniformly around it- The amplitude of the electric field of the light a distance of 1 m from the diode is -2-45 V-m6 V-m5-48 V-m1-73 V-m

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Toppr Admin · Accepted Answer

Sine we know that- intensity I-P4-x3C0-r2Also- I-12-x3B5-0E20-xD7-cP4-x3C0-r2-12-x3B5-0-xD7-E20-xD7-cE20-2P4-x3C0-x3B5-0r2c2-xD7-0-1-xD7-9-xD7-1091-xD7-3-xD7-108-6

A red LED emits light at $0.1$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $1 m$ from the diode is :
$2.45 V / m$
$6 V / m$
$5.48 V / m$
$1.73 V / m$

Sine we know that, intensity $I = \frac{P}{4 π r^{2}}$
Also, $I = \frac{1}{2} ε_{0} E_{0}^{2} \times c$
$\frac{P}{4 π r^{2}} = \frac{1}{2} ε_{0} \times E_{0}^{2} \times c$
$E_{0}^{2} = \frac{2 P}{4 π ε_{0} r^{2} c}$
$\frac{2 \times 0.1 \times 9 \times 10^{9}}{1 \times 3 \times 10^{8}} = 6$

A red LED emits light at0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1m from the diode is :2.45V/m6V/m5.48V/m1.73V/m

Sine we know that, intensity I=P4πr2Also, I=12ε0E20×cP4πr2=12ε0×E20×cE20=2P4πε0r2c2×0.1×9×1091×3×108=6

A red LED emits light at $0.1$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $1 m$ from the diode is :
$2.45 V / m$
$6 V / m$
$5.48 V / m$
$1.73 V / m$

Sine we know that, intensity $I = \frac{P}{4 π r^{2}}$
Also, $I = \frac{1}{2} ε_{0} E_{0}^{2} \times c$
$\frac{P}{4 π r^{2}} = \frac{1}{2} ε_{0} \times E_{0}^{2} \times c$
$E_{0}^{2} = \frac{2 P}{4 π ε_{0} r^{2} c}$
$\frac{2 \times 0.1 \times 9 \times 10^{9}}{1 \times 3 \times 10^{8}} = 6$