Question

question 2 (mandatory) (1 point) if the two waves of light depicted in the figure were compared, what could be determined about their frequencies? a) a has a 3.33×10^(-9) times lower frequency than b b) a has half the frequency of b. c) a has double the frequency of b. d) there is no relationship between the frequency and the wavelength of the two waves. e) a has a 3.0×10^(8) times higher frequency than b.

Explanation:

Step1: Recall wave - frequency relation

The speed of light $c$ is related to frequency $f$ and wavelength $\lambda$ by the formula $c = f\lambda$. Since the speed of light in a vacuum is constant ($c\approx3\times 10^{8}\ m/s$), $f=\frac{c}{\lambda}$.

Step2: Analyze wavelengths in figure

From the figure, we can see that the wavelength of wave A ($\lambda_A$) is twice the wavelength of wave B ($\lambda_B$), i.e., $\lambda_A = 2\lambda_B$.

Step3: Calculate frequency ratio

Let $f_A$ be the frequency of wave A and $f_B$ be the frequency of wave B. Using $f=\frac{c}{\lambda}$, we have $f_A=\frac{c}{\lambda_A}$ and $f_B=\frac{c}{\lambda_B}$. Substituting $\lambda_A = 2\lambda_B$ into the frequency - wavelength formula, we get $f_A=\frac{c}{2\lambda_B}$ and $f_B=\frac{c}{\lambda_B}$. So, $f_A=\frac{1}{2}f_B$.

Answer:

b) A has half the frequency of B.