A film with index of refraction 1.50 coats a glass lens with index of refraction 1.80. What is the minimum thickness of the thin film that will strongly reflect light with wavelength 600 nm?

Options

A. 150 nm
B. 200 nm
C. 300 nm
D. 450 nm

Explanation

To find the minimum thickness of the thin film that will strongly reflect light with a wavelength of 600 nm, we can follow these steps:

### Step 1: Understand the Condition for Strong Reflection
When light reflects off a thin film, constructive interference occurs when the path difference between the two reflected waves is an integer multiple of the wavelength. However, since the light reflects off a medium with a higher refractive index (from the film to the glass), there is a phase change of 180 degrees (or half a wavelength) for the wave reflecting from the top surface of the film.

### Step 2: Determine the Effective Wavelength in the Film
The wavelength of light in the film is given by the formula:
\[ \lambda' = \frac{\lambda}{n} \]
where:
- \(\lambda\) is the wavelength of light in vacuum (600 nm),
- \(n\) is the refractive index of the film (1.50).

Calculating the effective wavelength:
\[ \lambda' = \frac{600 \, \text{nm}}{1.50} = 400 \, \text{nm} \]

### Step 3: Set Up the Condition for Constructive Interference
For constructive interference in a thin film, the condition is given by:
\[ 2T = m \lambda' \]
where:
- \(T\) is the thickness of the film,
- \(m\) is the order of interference (an integer).

Since we want the minimum thickness, we set \(m = 1\):
\[ 2T = 1 \cdot \lambda' \]
Substituting the effective wavelength:
\[ 2T = 400 \, \text{nm} \]

### Step 4: Solve for the Thickness \(T\)
Now, we can solve for \(T\):
\[ T = \frac{400 \, \text{nm}}{2} = 200 \, \text{nm} \]

### Conclusion
The minimum thickness of the thin film that will strongly reflect light with a wavelength of 600 nm is:
\[ \boxed{200 \, \text{nm}} \]

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