A thin film with index of refraction 1.33 coats a glass lens with index of refraction 1.50. Which of the following choices is the smallest film thickness that will not reflect light with wavelength 640 nm?

Options

A. 160 nm
B. 240 nm
C. 360 nm
D. 480 nm

Explanation

To solve the problem of finding the smallest film thickness that will not reflect light with a wavelength of 640 nm, we will use the concept of destructive interference in thin films. Here’s a step-by-step solution:

### Step 1: Understand the Condition for Destructive Interference
For a thin film, destructive interference occurs when the path difference between the two reflected rays is equal to an odd multiple of half the wavelength. The condition can be expressed as:
\[ 2T = (m + \frac{1}{2}) \frac{\lambda}{n} \]
where:
- \( T \) = thickness of the film
- \( m \) = order of interference (0, 1, 2, ...)
- \( \lambda \) = wavelength of light in vacuum
- \( n \) = refractive index of the film

### Step 2: Substitute Known Values
In this case:
- \( \lambda = 640 \, \text{nm} \)
- \( n = 1.33 \) (the refractive index of the thin film)

### Step 3: Calculate for the Smallest Value of \( m \)
We will first calculate for \( m = 0 \):
\[ 2T = (0 + \frac{1}{2}) \frac{640 \, \text{nm}}{1.33} \]
\[ 2T = \frac{1}{2} \cdot \frac{640}{1.33} \]
\[ 2T = \frac{640}{2.66} \]
\[ T = \frac{640}{5.32} \]
\[ T \approx 120.0 \, \text{nm} \]

### Step 4: Check for Validity of the Thickness
Since 120 nm is not among the options provided, we will check for the next value of \( m \), which is \( m = 1 \):
\[ 2T = (1 + \frac{1}{2}) \frac{640 \, \text{nm}}{1.33} \]
\[ 2T = \frac{3}{2} \cdot \frac{640}{1.33} \]
\[ 2T = \frac{3 \times 640}{2.66} \]
\[ T = \frac{3 \times 640}{5.32} \]
\[ T \approx 360.0 \, \text{nm} \]

### Step 5: Conclusion
The smallest film thickness that will not reflect light with a wavelength of 640 nm is approximately **360 nm**.