A thin layer of oil (n = 1.25) covers the surface of a puddle of water (n = 1.33). If light of wavelength 500 nm in vacuum is minimally reflected (film appears dark) from the oil surface, what is the minimum non-zero thickness of the oil? 500 nm 400 nm 300 nm 200 nm 100 nm O

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### Optics in Thin Films

**Question 9 of 15**

A thin layer of oil (n = 1.25) covers the surface of a puddle of water (n = 1.33). If light of wavelength 500 nm in vacuum is minimally reflected (the film appears dark) from the oil surface, what is the minimum non-zero thickness of the oil?

- 500 nm
- 400 nm
- 300 nm
- 200 nm
- 100 nm

**Explanation:**
In this scenario, we are dealing with interference effects in thin films. When light reflects off the top and bottom surfaces of the oil layer, constructive and destructive interference can occur. For the film to appear dark (destructive interference), the path difference between the two reflected rays must be equal to half the wavelength of the light in the film plus any phase shift due to reflection. The minimum non-zero thickness of the oil can be found using the interference condition for destructive interference. 

Let's calculate the thickness using the formula for destructive interference in thin films: \( m + \frac{1}{2} = \frac{2nt}{\lambda} \)

Where:
- \( m \) is the order of interference (0, 1, 2, ...)
- \( n \) is the refractive index of the oil
- \( t \) is the thickness of the oil layer
- \( \lambda \) is the wavelength of light in vacuum

The smallest non-zero thickness corresponds to \( m = 0 \). Solving for \( t \):

\[ \frac{1}{2} = \frac{2 \cdot 1.25 \cdot t}{500 \, \text{nm}} \]
\[ t = \frac{500 \, \text{nm}}{4 \cdot 1.25} \]
\[ t = 100 \, \text{nm} \]

Thus, the minimum non-zero thickness of the oil for destructive interference of 500 nm light is **100 nm**. 

**Answer:** 100 nm
Transcribed Image Text:### Optics in Thin Films **Question 9 of 15** A thin layer of oil (n = 1.25) covers the surface of a puddle of water (n = 1.33). If light of wavelength 500 nm in vacuum is minimally reflected (the film appears dark) from the oil surface, what is the minimum non-zero thickness of the oil? - 500 nm - 400 nm - 300 nm - 200 nm - 100 nm **Explanation:** In this scenario, we are dealing with interference effects in thin films. When light reflects off the top and bottom surfaces of the oil layer, constructive and destructive interference can occur. For the film to appear dark (destructive interference), the path difference between the two reflected rays must be equal to half the wavelength of the light in the film plus any phase shift due to reflection. The minimum non-zero thickness of the oil can be found using the interference condition for destructive interference. Let's calculate the thickness using the formula for destructive interference in thin films: \( m + \frac{1}{2} = \frac{2nt}{\lambda} \) Where: - \( m \) is the order of interference (0, 1, 2, ...) - \( n \) is the refractive index of the oil - \( t \) is the thickness of the oil layer - \( \lambda \) is the wavelength of light in vacuum The smallest non-zero thickness corresponds to \( m = 0 \). Solving for \( t \): \[ \frac{1}{2} = \frac{2 \cdot 1.25 \cdot t}{500 \, \text{nm}} \] \[ t = \frac{500 \, \text{nm}}{4 \cdot 1.25} \] \[ t = 100 \, \text{nm} \] Thus, the minimum non-zero thickness of the oil for destructive interference of 500 nm light is **100 nm**. **Answer:** 100 nm
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