A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is [2023]

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Solution

Q.
A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is [2023]

Solution

Q.
A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is [2023]

1 $\frac{1}{2 π} \frac{W}{m^{2}}$

2 $\frac{1}{40 π} \frac{W}{m^{2}}$

3 $\frac{1}{10 π} \frac{W}{m^{2}}$

4 $\frac{1}{20 π} \frac{W}{m^{2}}$

Ans.
(2)

$I_{E F} = \frac{1}{2} \times \frac{5}{4 π \times 5^{2}} = \frac{1}{40 π} {W/m}^{2}$

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Solution

Q. A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is [2023]

1 12πWm2

2 140πWm2

3 110πWm2

4 120πWm2

Ans. (2) IEF=12×54π×52=140πW/m2

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Q.
A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is [2023]

1 $\frac{1}{2 π} \frac{W}{m^{2}}$

2 $\frac{1}{40 π} \frac{W}{m^{2}}$

3 $\frac{1}{10 π} \frac{W}{m^{2}}$

4 $\frac{1}{20 π} \frac{W}{m^{2}}$

Ans.
(2)

$I_{E F} = \frac{1}{2} \times \frac{5}{4 π \times 5^{2}} = \frac{1}{40 π} {W/m}^{2}$