A 600 nm light is perpendicularly incident on a soap film suspended air. The film is `1.00 mu m` thick with `n = 1.35`. Which statement most accurately describes the interference of light reflected by the two surfaces of the film?

Options

A. The waves are close to destructive interference.
B. The waves are closed to constructive interference.
C. The waves how complete destructive interference.
D. The waves how complete constructive interference.

Explanation

To solve the problem, we need to analyze the interference of light reflected from the two surfaces of a soap film. Here’s a step-by-step breakdown:

### Step 1: Identify the given parameters
- Wavelength of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Thickness of the soap film, \( T = 1.00 \, \mu m = 1.00 \times 10^{-6} \, \text{m} \)
- Refractive index of the soap film, \( n = 1.35 \)

### Step 2: Determine the phase change upon reflection
When light reflects off a boundary from a medium of lower refractive index to a medium of higher refractive index (air to soap film), a phase change of \( \pi \) (or half a wavelength) occurs. In this case, the first reflection (from air to soap) does not have a phase change, while the second reflection (from soap to air) does.

### Step 3: Calculate the total path difference
The total path difference \( \Delta x \) for light reflecting off both surfaces of the film is given by:
\[ \Delta x = 2 \times T \times n \]
Substituting the values:
\[ \Delta x = 2 \times (1.00 \times 10^{-6} \, \text{m}) \times 1.35 = 2.70 \times 10^{-6} \, \text{m} \]

### Step 4: Convert the path difference to wavelength units
To find the number of wavelengths in the path difference, we convert the path difference from meters to nanometers:
\[ \Delta x = 2.70 \times 10^{-6} \, \text{m} = 2.70 \times 10^{3} \, \text{nm} \]
Now, we calculate how many wavelengths fit into this path difference:
\[ \text{Number of wavelengths} = \frac{\Delta x}{\lambda} = \frac{2.70 \times 10^{3} \, \text{nm}}{600 \, \text{nm}} = 4.5 \]

### Step 5: Determine the type of interference
Since there is a \( \pi \) phase change upon reflection from the second surface, we need to consider this in our calculations. The condition for constructive interference is:
\[ \Delta x = (m + \frac{1}{2}) \lambda \quad \text{(with phase change)} \]
For destructive interference, it is:
\[ \Delta x = m \lambda \quad \text{(without phase change)} \]

Since \( 4.5 \) is not an integer, it indicates that the interference is destructive. However, because of the phase change, we can conclude that the condition for constructive interference is satisfied when \( m = 4 \).

### Conclusion
Thus, the interference of light reflected by the two surfaces of the film results in complete constructive interference.

### Final Answer
The statement that most accurately describes the interference of light reflected by the two surfaces of the film is that the waves show complete constructive interference.

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