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This problem deals with interference in thin films, where the fiber acts as a spacer between two flat glass plates, creating an interference pattern when illuminated with light. 

To find the diameter \( d \) of the fiber, we can use the following relationship:

\[
d = \frac{\lambda L}{2s}
\]

where:
- \( \lambda \) is the wavelength of the light used (\( 640 \, \text{nm} = 640 \times 10^{-9} \, \text{m} \)),
- \( L \) is the length of the plates (\( 15.5 \, \text{cm} = 0.155 \, \text{m} \)),
- \( s \) is the distance between the interference bands (\( 0.535 \, \text{mm} = 0.535 \times 10^{-3} \, \text{m} \)).

Let’s plug in these values to find \( d \).

\[
d = \frac{(640 \times 10^{-9}) \times 0.155}{2 \times (0.535 \times 10^{-3})}
\]

Now, calculating:

\[
d = \frac{9.92 \times 10^{-8}}{1.07 \times 10^{-3}} = 9.27 \times 10^{-5} \, \text{m}
\]

Converting this to micrometers (1 m = \( 10^6 \) µm):

\[
d = 9.27 \times 10^{-5} \times 10^6 = 92.7 \, \mu m
\]

So, the diameter of the fiber is approximately **92.7 µm**.

Would you like a deeper explanation of the interference concept used here or have any questions about the process?

Here are 5 related questions to explore:
1. How does the wavelength of light affect the interference pattern?
2. What would happen to the interference pattern if the plates were tilted?
3. Why do we use interference to measure such small distances?
4. What assumptions are made when applying this equation to real-world measurements?
5. How can the accuracy of this measurement be improved in a lab setting?

**Tip**: When converting between units, always ensure consistency (e.g., all measurements in meters or micrometers) to avoid errors in the final result.

An investigator finds a fiber at a crime scene that he wishes to use as evidence against a suspect. The technician places it between two flat glass plates illuminated with light of wavelength 640 nm, observing interference bands separated by 0.535 mm. What is the diameter of the fiber?

Learn how to calculate the diameter of a fiber using interference patterns. This problem involves advanced physics concepts, including the interference pattern theorem and unit conversions for wavelengths and distances. Suitable for advanced high school physics students.

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