Suppose that for an interferometer to work properly, the electromagnetic waves must arrive in sync to 1% of the period of the wave. What time difference is this for the radio wave of wavelength 42 cm? What time difference is this for a 600-nm visible wavelength wave? (Recall that the period of the wave is the inverse of its frequency v.)

The time difference for the 42-cm wavelength is    × 10−11s.

The time difference for the 600-nm wavelength is    × 10−17s.

To determine the time difference for the radio wave of wavelength 42 cm and the visible wavelength of 600 nm for the interferometer, we first need to calculate the period of each wave. The period \( T \) is the inverse of the frequency \( v \), and the frequency can be derived from the speed of light \( c \) and the wavelength \( \lambda \).

The speed of light \( c \) is approximately \( 3 \times 10^8 \) meters per second.

1. **Calculate the period for the 42 cm wavelength:**

   - Wavelength \( \lambda = 42 \) cm \( = 0.42 \) meters.
   - The frequency \( v = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{0.42 \, \text{m}} \).

     \[
     v = \frac{3 \times 10^8}{0.42} \approx 7.14 \times 10^8 \, \text{Hz}
     \]

   - The period \( T = \frac{1}{v} \).

     \[
     T = \frac{1}{7.14 \times 10^8} \approx 1.4 \times 10^{-9} \, \text{s}
     \]

   - To sync to 1% of the period:

     \[
     \text{Time difference} = 0.01 \times T = 0.01 \times 1.4 \times 10^{-9} \approx 1.4 \times 10^{-11} \, \text{s}
     \]

   Therefore, the time difference for the 42-cm wavelength is \( 1.4 \times 10^{-11} \) s.

2. **Calculate the period for the 600 nm wavelength:**

   - Wavelength \( \lambda = 600 \) nm \( = 600 \times 10^{-9} \) meters.
   - The frequency \( v = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{600 \times 10^{-9} \, \text{m}} \).

     \[
     v = \frac{3 \times 10^8}{600 \times 10^{-9}} = 5 \times 10^{14} \, \text{Hz}
     \]

   - The period \( T = \frac{1}{v} \).

     \[
     T = \frac{1}{5 \times 10^{14}} = 2 \times 10^{-15} \, \text{s}
     \]

   - To sync to 1% of the period:

     \[
     \text{Time difference} = 0.01 \times T = 0.01 \times 2 \times 10^{-15} = 2 \times 10^{-17} \, \text{s}
     \]

   Therefore, the time difference for the 600-nm wavelength is \( 2 \times 10^{-17} \) s.

Would you like more details or have any questions?

Here are some questions you might consider next:

1. What is the relationship between wavelength and frequency for electromagnetic waves?
2. How does the speed of light influence the period of a wave?
3. Why is synchronizing electromagnetic waves important for interferometers?
4. Can you explain how the time difference impacts the accuracy of an interferometer?
5. How do different wavelengths of light affect interferometry experiments?

**Tip:** Understanding the relationship between wavelength, frequency, and the speed of light is crucial for solving problems related to electromagnetic waves.

Explore how to calculate the time differences required for interferometers using electromagnetic waves of different wavelengths. This includes a detailed calculation for a 42 cm radio wave and a 600 nm visible light wave.

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