Question

Assume a bulb of efficiency 2.5% as a point source. The peak values of electric filed produced by the radiation coming from a 100 w bulb at a distance of 3 m is respectively:-

• A. $2.5V{m}^{-1}$
• B. $4.2V{m}^{-1}$
• C. $4.07V{m}^{-1}$
• D. $3.6V{m}^{-1}$

Answer 1

The correct option is C $4.07V{m}^{-1}$

The bulb as a point source ,radiates light in all directions uniformly.

At a distance of $3m$, the surface area of the surrounding sphere is

$A=4\pi r^2=4\pi {\left(3\right)}^2=11m^2$

Then intensity at this distance is

$I=\frac{p}{a}=\frac{100\times 2.5}{113}=0.022\frac{W}{m^2}$

Half of this intensity is provided by the electric field and half by the magnetic field

$\begin{array}{c}=\frac{1}{2}\left(0.022\frac{W}{m^2}\right)=\frac{1}{2}\left({\in }_0{E}_{rms}^{2}c\right)\\ E_{rms}=\sqrt{\left[\frac{\left(0.222\right)}{\left(8.85\times {10}^{-12}\times 3\times {10}^8\right)}\right]\frac{V}{m}}=2.9\frac{V}{m}\\ E_0=\sqrt{2}{E}_{rms}=\sqrt{2}\times 2.9V=4.07\frac{V}{m}\end{array}$

Option $C$ is correct .