In a Michelson-type experiment, a rotating eight-sided mirror was placed 50.0 km from the reflecting mirror. The observer found that in order to observe the return light ray, the mirror had to rotate at 375 Hz. What is the speed of light calculated from this experiment

In a Michelson-type experiment, the speed of light \( c \) is calculated using the known rotation frequency of the rotating mirror, the number of sides of the mirror, and the distance between the rotating mirror and the reflecting mirror.

Let's summarize the known quantities:
- The distance between the rotating mirror and the reflecting mirror: \( d = 50.0 \, \text{km} = 50.0 \times 10^3 \, \text{m} \).
- The rotating mirror has 8 sides.
- The rotation frequency of the mirror is \( f = 375 \, \text{Hz} \).

The light travels from the rotating mirror to the reflecting mirror and back, so the total distance traveled by the light is \( 2d \).

For the observer to see the light, the rotating mirror must rotate by one-eighth of a full rotation (because the mirror has 8 sides) during the time it takes the light to travel that distance. Therefore, the time for the light to make a round trip must match the time it takes the mirror to rotate by \( \frac{1}{8} \) of a full rotation.

The time \( t \) for the mirror to rotate by \( \frac{1}{8} \) of a full rotation is:

\[
t = \frac{1}{8f}
\]

Since the speed of light \( c \) is the distance traveled divided by the time, and the light travels a distance \( 2d \) in time \( t \), we can write:

\[
c = \frac{2d}{t} = \frac{2d}{\frac{1}{8f}} = 16df
\]

Now, substitute the known values:

\[
c = 16 \times (50.0 \times 10^3 \, \text{m}) \times 375 \, \text{Hz}
\]

\[
c = 16 \times 50.0 \times 10^3 \times 375
\]

\[
c = 300 \times 10^6 \, \text{m/s}
\]

Therefore, the speed of light calculated from this experiment is \( c = 3.00 \times 10^8 \, \text{m/s} \).

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**Related Questions:**
1. How does the number of sides of the rotating mirror affect the result?
2. What would happen if the frequency were increased in this experiment?
3. Why is the return trip of light significant in this experiment?
4. How does the Michelson-type experiment compare to modern methods for measuring the speed of light?
5. What assumptions are made about the reflectivity and alignment of the mirrors?

**Tip:** Precision in measuring both the frequency and distance is crucial for accurate results in optical experiments like these.

In a Michelson-type experiment, a rotating eight-sided mirror was placed 50.0 km from the reflecting mirror. The observer found that in order to observe the return light ray, the mirror had to rotate at 375 Hz. What is the speed of light calculated from this experiment?

Learn how to calculate the speed of light using a Michelson-type experiment with an eight-sided rotating mirror placed 50.0 km from a reflecting mirror. In this problem, the mirror rotates at 375 Hz, and the derived formula c = 16df is used to calculate the speed of light. Suitable for undergraduate physics students.

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