Question:
A laser beam of wavelength $1\text{ }\mu\text{m}$ is split and sent into two vacuum cavities of equal length $L$ as shown in the figure. A detector can register an interference signal only if the phase difference between the returning beams is at least $5 \times 10^{-11}\text{ rad}$. A certain physical effect changes the length of cavity 2 by an amount $\Delta L$ such that $\Delta L / L \approx 10^{-21}$. The minimum cavity length (in km) needed for measuring this physical effect is approximately
A laser beam of wavelength $1\text{ }\mu\text{m}$ is split and sent into two vacuum cavities of equal length $L$ as shown in the figure. A detector can register an interference signal only if the phase difference between the returning beams is at least $5 \times 10^{-11}\text{ rad}$. A certain physical effect changes the length of cavity 2 by an amount $\Delta L$ such that $\Delta L / L \approx 10^{-21}$. The minimum cavity length (in km) needed for measuring this physical effect is approximately
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Remember that in Michelson-type interferometers or cavities, the light travels a round trip, so the change in path length is $2\Delta L$ instead of just $\Delta L$.
Updated On: Jun 16, 2026
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Solution and Explanation
Step 1: Understanding the Question:
The problem describes an interferometer with two cavities of length $L$.
A physical change alters the length of one cavity, creating a phase difference between the two returning beams. We need to find the minimum length $L$ required to detect this change.
Step 2: Key Formula or Approach:
- The phase difference $\Delta \phi$ caused by a path length difference $\Delta x$ is:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- Since the laser beams travel back and forth in the cavities, the path difference is:
\[ \Delta x = 2 \Delta L \]
Step 3: Detailed Explanation:
• The wavelength of the laser is $\lambda = 1\text{ }\mu\text{m} = 10^{-6}\text{ m}$.
• The minimum detectable phase difference is $\Delta \phi_{\min} = 5 \times 10^{-11}\text{ rad}$.
• The relative length change is:
\[ \frac{\Delta L}{L} = 10^{-21} \implies \Delta L = 10^{-21} L \]
• The phase difference created by the round-trip change of length in cavity 2 is:
\[ \Delta \phi = \frac{4\pi \Delta L}{\lambda} \]
• To detect this signal, we need:
\[ \Delta \phi \ge \Delta \phi_{\min} \]
\[ \frac{4\pi (10^{-21} L)}{\lambda} \ge 5 \times 10^{-11} \]
• Substituting $\lambda = 10^{-6}\text{ m}$:
\[ \frac{4\pi \cdot 10^{-21} L}{10^{-6}} \ge 5 \times 10^{-11} \]
\[ 4\pi \cdot 10^{-15} L \ge 5 \times 10^{-11} \]
\[ L \ge \frac{5 \times 10^{-11}}{4\pi \cdot 10^{-15}} \]
\[ L \ge \frac{50000}{4\pi}\text{ m} \approx 3978\text{ m} \approx 4\text{ km} \]
Step 4: Final Answer:
The minimum cavity length needed is approximately 4 km.
The problem describes an interferometer with two cavities of length $L$.
A physical change alters the length of one cavity, creating a phase difference between the two returning beams. We need to find the minimum length $L$ required to detect this change.
Step 2: Key Formula or Approach:
- The phase difference $\Delta \phi$ caused by a path length difference $\Delta x$ is:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- Since the laser beams travel back and forth in the cavities, the path difference is:
\[ \Delta x = 2 \Delta L \]
Step 3: Detailed Explanation:
• The wavelength of the laser is $\lambda = 1\text{ }\mu\text{m} = 10^{-6}\text{ m}$.
• The minimum detectable phase difference is $\Delta \phi_{\min} = 5 \times 10^{-11}\text{ rad}$.
• The relative length change is:
\[ \frac{\Delta L}{L} = 10^{-21} \implies \Delta L = 10^{-21} L \]
• The phase difference created by the round-trip change of length in cavity 2 is:
\[ \Delta \phi = \frac{4\pi \Delta L}{\lambda} \]
• To detect this signal, we need:
\[ \Delta \phi \ge \Delta \phi_{\min} \]
\[ \frac{4\pi (10^{-21} L)}{\lambda} \ge 5 \times 10^{-11} \]
• Substituting $\lambda = 10^{-6}\text{ m}$:
\[ \frac{4\pi \cdot 10^{-21} L}{10^{-6}} \ge 5 \times 10^{-11} \]
\[ 4\pi \cdot 10^{-15} L \ge 5 \times 10^{-11} \]
\[ L \ge \frac{5 \times 10^{-11}}{4\pi \cdot 10^{-15}} \]
\[ L \ge \frac{50000}{4\pi}\text{ m} \approx 3978\text{ m} \approx 4\text{ km} \]
Step 4: Final Answer:
The minimum cavity length needed is approximately 4 km.
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