NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

1 Medium

1 of 1

Application +4 | -1

The amplitude of electric field in a parallel beam of light of intensity

4\mathrm{\;W}{\mathrm{m}^{ - 2}}

View answer and explanation

Correct option: D

55.5\mathrm{\;N}{\mathrm{C}^{ - 1}}

Option D (

55.5\mathrm{\;N}{\mathrm{C}^{ - 1}}

) is correct because the amplitude of the electric field is calculated using the intensity formula

I = \frac{1}{2}c\epsilon_0 E_0^2

More Questions from ELECTROMAGNETIC WAVES

Practice All Chapter MCQs →

{E_X}

{E_R}

{E_M}

60\,\,\frac{V}{m}

{E_1} = {E_{01}}\hat x\cos \left[ {2\pi v\left( {\frac{z}{c} - t} \right)} \right]

{E_2} = {E_{02}}\hat x\cos \left[ {k\left( {2z - ct} \right)} \right]

{\varepsilon _{{r_1}}}

{\varepsilon _{{r_2}}}

E = 50\sin \,(500x - 10 \times {10^{10}}t)\,v/m.

c

\mathrm{\varepsilon }

\mathrm{\mu }

\sqrt {\frac{\mathrm{\mu }}{\mathrm{\varepsilon }}}