Question

The relationship between wavelength (\( \lambda \)), frequency (\( \nu \)), and velocity (\( c \)) of electromagnetic wave is

• A. \( c = \frac{\nu^2}{\lambda} \)
• B. \( c = \frac{\nu}{\lambda} \)
• C. \( c = \nu \lambda \)
• D. \( c = \nu \lambda^2 \)

Hint:

Remember, the velocity of an electromagnetic wave is the product of its frequency and wavelength, as given by the equation \( c = \nu \lambda \). This is a key relationship in wave motion.

Answer 1

The Correct Option is C

To solve this problem, we need to apply the fundamental relationship between the velocity, frequency, and wavelength of electromagnetic waves.
The velocity \( c \) of an electromagnetic wave is the speed at which the wave propagates through space. The frequency \( \nu \) is the number of oscillations or cycles of the wave that pass a point per second, and the wavelength \( \lambda \) is the distance between two consecutive points in phase, such as two peaks of the wave.
From classical wave theory, we know that the velocity \( c \), frequency \( \nu \), and wavelength \( \lambda \) of any electromagnetic wave are related by the equation:
\[ c = \nu \lambda \] This means that the velocity of a wave is equal to the product of its frequency and its wavelength. This relationship is true for all types of electromagnetic waves, including light, radio waves, and X-rays.
Let's analyze the options:
- Option (A): \( c = \frac{\nu^2}{\lambda} \) is incorrect because it suggests that the velocity is proportional to the square of the frequency, which is not correct.
- Option (B): \( c = \frac{\nu}{\lambda} \) is also incorrect because it suggests that velocity is the ratio of frequency to wavelength, which contradicts the standard formula.
- Option (C): \( c = \nu \lambda \) is correct. This is the fundamental equation that links velocity, frequency, and wavelength.
- Option (D): \( c = \nu \lambda^2 \) is incorrect because the square of the wavelength does not appear in the correct formula.
Thus, the correct answer is (C), \( c = \nu \lambda \).