A thin uniform film of refractive index 1.75 is placed on a sheet of glass of refractive index 1.5. At room temperature `(20^(@) C)` this film is just thick enoungh for light with wavelength 600 nm reflected off the top of the film to be canceled by light reflected from the top the glass. After the glass is placed in on oven and slowly heated to `170^(@) C`, the film conceals reflected light with wavelength 606 nm. The coefficient of linear expansion of the film is (ignore any changes in the refractive index of the film due to the temperature change)

Options

A. `3.3 xx 10^(-5).^@C^(-1)`
B. `6.6 xx 10^(-5).^@C^(-1)`
C. `9.9 xx 10^(-5).^@C^(-1)`
D. `2.2 xx 10^(-5).^@C^(-1)`

Explanation

To solve the problem, we need to analyze the conditions for destructive interference in the thin film and how the change in temperature affects the thickness of the film and the wavelength of light reflected.

### Step-by-Step Solution:

1. **Understanding Destructive Interference**:
For destructive interference in a thin film, the condition is given by:
\[ 2t = (m + \frac{1}{2}) \lambda/n \]
where \( t \) is the thickness of the film, \( m \) is an integer (order of interference), \( \lambda \) is the wavelength of light in vacuum, and \( n \) is the refractive index of the film.

2. **Initial Condition at Room Temperature**:
At room temperature (20°C), the wavelength of light is \( \lambda_1 = 600 \, \text{nm} \) and the refractive index of the film is \( n = 1.75 \). The thickness \( t_1 \) can be expressed as:
\[ 2t_1 = (m + \frac{1}{2}) \frac{\lambda_1}{n} \]
Rearranging gives:
\[ t_1 = \frac{(m + \frac{1}{2}) \lambda_1}{2n} \]

3. **Condition at Elevated Temperature**:
At 170°C, the wavelength of light changes to \( \lambda_2 = 606 \, \text{nm} \). The new thickness \( t_2 \) can be expressed similarly:
\[ 2t_2 = (m + \frac{1}{2}) \frac{\lambda_2}{n} \]
Rearranging gives:
\[ t_2 = \frac{(m + \frac{1}{2}) \lambda_2}{2n} \]

4. **Relating Thickness and Temperature**:
The thickness of the film changes with temperature according to the linear expansion formula:
\[ t_2 = t_1(1 + \alpha \Delta T) \]
where \( \alpha \) is the coefficient of linear expansion and \( \Delta T = 170 - 20 = 150 \, \text{°C} \).

5. **Equating Thicknesses**:
From the expressions for \( t_1 \) and \( t_2 \):
\[ \frac{(m + \frac{1}{2}) \lambda_2}{2n} = \frac{(m + \frac{1}{2}) \lambda_1}{2n}(1 + \alpha \Delta T) \]
Canceling common terms gives:
\[ \lambda_2 = \lambda_1(1 + \alpha \Delta T) \]

6. **Substituting Values**:
Substituting \( \lambda_1 = 600 \, \text{nm} \), \( \lambda_2 = 606 \, \text{nm} \), and \( \Delta T = 150 \):
\[ 606 = 600(1 + \alpha \cdot 150) \]
Rearranging gives:
\[ 1 + \alpha \cdot 150 = \frac{606}{600} = 1.01 \]
Therefore:
\[ \alpha \cdot 150 = 0.01 \]
\[ \alpha = \frac{0.01}{150} = \frac{1}{15000} \approx 6.67 \times 10^{-5} \, \text{°C}^{-1} \]

### Final Answer:
The coefficient of linear expansion of the film is approximately:
\[ \alpha \approx 6.67 \times 10^{-5} \, \text{°C}^{-1} \]