Question

The poynting vector of an electromagnetic wave in vacuum is s= 120w/m^2)sin^2(8.0rad/m)z+(2.4×10^9rad/s)t)k.what is the wavelngth?

Answer 1

we know that the standard eq of electromagnetic wave is,

E = E₀sin(kx-wt)i

The given equation is

s= 120w/m^2)sin^2(8.0rad/m)z+(2.4×10^9rad/s)t)k

Comparing both the equation, we get

k = 8 rad/m

we know that wavelength λ = 2π/k

=> λ = 2*3.14/8

= 0.3925 m

Hence the wavelength of the wave is 0.3925 m

Answer 2

Given:

Poynting vector, s = ( 120 w / m^2 ) sin^2 ( 8.0 rad / m ) z + ( 2.4 * 10^9 rad / s  ) t ) k

To find:

The wavelength.

Solution:

The equation of the electromagnetic wave,

E = E₀ sin^2 ( k x - w t ) i

Comparing,

We get,

k = 8

Hence,

Wavelength = 2 * π / k

π - 3.14

Substituting,

We get,

2 * 3.14 / 8

Hence, Wavelength = 0.3925 m

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