JEE PHYSICS MCQ: WAVES

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The electric field part of an electromagnetic wave in vacuum is

\mathrm{E}=3.1 \mathrm{~N~C}^{-1} \times \cos \left[\left(1.8 \mathrm{~rad} \mathrm{~m}^{-1}\right) \mathrm{y}+\left(5.4 \times 10^8 \mathrm{~rad} \mathrm{~s}^{-1}\right) \mathrm{t}\right] \hat{\mathrm{i}}

The wavelength of this part of electromagnetic wave is

View answer and explanation

Correct option: D

3.5 \mathrm{~m}

Option D (

3.5 \mathrm{~m}

) is correct because it is the wavelength calculated from the wave number (k) in the given electromagnetic wave equation.

More Questions from WAVES

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3\,\,kHz

0.5

f\mathrm{ }Hz

V\mathrm{ }m/sec

0.2\,\,V

y = \frac{3}{{{{\left( {2x + 3t} \right)}^2} + 4}},

(– \mathrm{ve}) \,x

y = f(kx +\omega t),

X

0.01\,\,s

5 \mathrm{~m}

245 \mathrm{~g}

1 \mathrm{~kg}

g=980 \mathrm{~cm} \mathrm{~s}^{-2}

T

V