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(A) From the given figure, the distance shown corresponds to one-fourth of the wavelength (from crest to node), i.e., $\frac{\lambda}{4} = 1.5 \ pm$.T

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) From the given figure, the distance shown corresponds to one-fourth of the wavelength (from crest to node), i.e., $\frac{\lambda}{4} = 1.5 \ pm$.Therefore, the wavelength $\lambda = 4 	imes 1.5 \ pm = 6 \ pm = 6 	imes 10^{-12} \ m$.The relationship between frequency $(
u)$, wavelength $(\lambda)$, and speed of light $(c)$ is $
u = \frac{c}{\lambda}$.Given $c = 3 	imes 10^8 \ m/s$, we have $
u = \frac{3 	imes 10^8 \ m/s}{6 	imes 10^{-12} \ m} = 0.5 	imes 10^{20} \ Hz$.Expressing this in the form $x 	imes 10^{19} \ Hz$, we get $5 	imes 10^{19} \ Hz$.Thus, $x = 5$.

$A$ hypothetical electromagnetic wave is shown below. The frequency of the wave is $x \times 10^{19} \ Hz$. $x =$ . . . . (nearest integer)

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