A 100 W electric bulb produces electromagnetic radiation with electric field amplitude of `(2V)/(m)` at a distance of 10 m.
Assuming it as a point source, estimate the efficiency of the bulb. 

Options

A. 0.049
B. 2.5%
C. 0.066
D. 0.197

Answer

C

Explanation

Intensity of radiations, `I=("Power")/("Area")xx" Efficiency"`
`=(100)/(4xx pi xx(10)^(3))xx eta`
Now, half of intensity is priovided by electric field and another half by magnetic field.
So, `(1)/(2)I=(1)/(2)epsilon_(0)E^(2)_("rns")c=(1)/(2)epsilon_(0)(E_("max")^(2))c`
`[E_("rns")=E_("max")//sqrt(2)]`
`implies" "(100)/(4pi xx10^(3))eta=(1)/(2)xx8.85xx10^(-12)xx(4)/(2)xx3xx10^(8)`
`implies" "eta=(8.85xx2xx3xx1256xx10^(-4))/(100)`
`implies" "eta=(6.66)/(100)`
or `" "eta=6.66%`