For an electromagnetic wave traveling in vacuum, what happens to the frequency $f$ as the wavelength $\lambda$ increases, given $c=f\lambda$?

• A. $f$ stays the same because $c$ is constant.
• B. $f$ becomes zero when $\lambda$ increases.
• C. $f$ decreases (it is inversely proportional to $\lambda$).
• D. $f$ increases (it is directly proportional to $\lambda$).

Answer: (C)

Verified step by step guidance

1. Recall the fundamental relationship for an electromagnetic wave traveling in vacuum: \(c = f \times \lambda\), where \(c\) is the speed of light, \(f\) is the frequency, and \(\lambda\) is the wavelength.
2. Understand that the speed of light \(c\) is a constant value in vacuum, approximately \$3 \times 10^{8}$ meters per second, and does not change regardless of the wavelength or frequency.
3. Since \(c\) is constant, any change in wavelength \(\lambda\) must be accompanied by a corresponding change in frequency \(f\) to keep the product \(f \times \lambda\) equal to \(c\).
4. Rearrange the equation to express frequency in terms of wavelength: \(f = \frac{c}{\lambda}\). This shows that frequency is inversely proportional to wavelength.
5. Therefore, as the wavelength \(\lambda\) increases, the frequency \(f\) decreases, because their product must remain equal to the constant speed of light \(c\).