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Question: The electric fields part of an electromagnetic wave I vacuum is E=3.1 \(\mathrm { C } ^ { - 1 }\) \(...

The electric fields part of an electromagnetic wave I vacuum is E=3.1 C−1\mathrm { C } ^ { - 1 } cos⁡[(1.8radm−1)y+(5.4×188rads−1)t]i^\cos\left\lbrack (1.8radm^{- 1})y + (5.4 \times 18^{8}rads^{- 1})t \right\rbrack\widehat{i}the wavelength of this part of electromagnetic wave is :

A

1.5m

B

2m

C

2.5m

D

3.5m

Answer

3.5m

Explanation

Solution

: Given E=3.1NC−1×E = 3.1NC^{- 1} \times

cos⁡[(1.8radm−1)y++(5.4×108rads−1)t]i^\cos\lbrack(1.8radm^{- 1})y + + (5.4 \times 10^{8}rads^{- 1})t\rbrack\widehat{i} …(i)

Camparing (i) with the equation

E=E0cos⁡(ky+ωt)E = E_{0}\cos(ky + \omega t) …(ii)

We get, k=1.8radm−1,k = 1.8radm^{- 1},

E0=3.1NC−1,c=3×108ms−1,ω=5.4×108rads−1E_{0} = 3.1 NC^{- 1},c = 3 \times 10^{8}ms^{- 1},\omega = 5.4 \times 10^{8}rads^{- 1}

Now, ∴υmυm=1011105=106\therefore\frac{\upsilon_{m}}{\upsilon_{m}} = \frac{10^{11}}{10^{5}} = 10^{6}