Question

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

Answer 1

Answer:

Explanation:

Light energy emitted per unit time

=

100

×

5

100

=

5

W

intensity at 5M

=

power

4

π

(

5

)

2

=

5

4

π

(

5

)

2

=

1

20

π

However,

1

20

π

is due to Both E and B with equal Contributions.

So, intensity due to

E

=

1

40

π

W

m

2