Question

A wave travels from one medium to another. Its frequency stays the same. If the speed in the first medium is greater than the speed in the second medium, how do the wavelengths compare?(1 point)
Responses

The wavelength in the first medium is shorter than the wavelength in the second medium.
The wavelength in the first medium is shorter than the wavelength in the second medium.

The wavelength in the first medium is longer than the wavelength in the second medium.
The wavelength in the first medium is longer than the wavelength in the second medium.

The wavelengths are the same because the frequency stays the same.
The wavelengths are the same because the frequency stays the same.

The wavelengths are the same because wavelengths can never change.

Answer 1

To analyze how the wavelengths compare when a wave travels from one medium to another, we can use the wave equation:

\[ v = f \lambda \]

where:
- \( v \) is the speed of the wave,
- \( f \) is the frequency, and
- \( \lambda \) is the wavelength.

Given that the frequency (\( f \)) stays the same as the wave transitions from one medium to another, and that the speed in the first medium (\( v_1 \)) is greater than the speed in the second medium (\( v_2 \)), we can express the wavelengths in both media:

1. For the first medium:
\[ v_1 = f \lambda_1 \]
Thus,
\[ \lambda_1 = \frac{v_1}{f} \]

2. For the second medium:
\[ v_2 = f \lambda_2 \]
Thus,
\[ \lambda_2 = \frac{v_2}{f} \]

Since \( v_1 > v_2 \), it follows that:
\[ \lambda_1 = \frac{v_1}{f} > \frac{v_2}{f} = \lambda_2 \]

Therefore, the wavelength in the first medium is longer than the wavelength in the second medium.

The correct response is:
**The wavelength in the first medium is longer than the wavelength in the second medium.**