Answer:
λ = 157 μm
Explanation:
given,
[tex]\vec{E}(y,t)=[ 3.9\times 10^5\ V/m cos(ky-(1.20\times 10^{13}\ rad/s )t)]\vec{k}[/tex]
now,
ω = 1.2 x 10¹³ rad/s
[tex]f =\dfrac{\omega}{2\pi}[/tex]
[tex]f =\dfrac{1.2\times 10^{13}}{2\pi}[/tex]
f = 1.91 x 10¹² Hz
determine the wavelength of the E M wave
[tex]\lambda = \dfrac{c}{f}[/tex]
[tex]\lambda = \dfrac{3\times 10^8}{1.91\times 10^{12}}[/tex]
λ = 1.57 x 10⁻⁴ m
λ = 157 x 10⁻⁶ m
λ = 157 μm
hence, the wavelength of the wave is equal to λ = 157 μm
Given that,
here,
As we know,
→ [tex]f = \frac{\omega}{2 \pi}[/tex]
By substituting the values, we get
[tex]= \frac{1.2\times 10 ^{13}}{2 \pi}[/tex]
[tex]= 1.91\times 10^{12} \ Hz[/tex]
hence,
The wavelength will be:
→ [tex]\lambda = \frac{c}{f}[/tex]
[tex]= \frac{3\times 10^8}{1.91\times 10^{12}}[/tex]
[tex]=157\times 10^{-6} \ m[/tex]
[tex]= 157 \ \mu m[/tex]
Thus the above response is correct.
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