A thin oil film of refracting index `1.2` floats on the surface of water `(mu=4/3)`. When a light of wavelength `lamda=9.6xx10^(-7)m` falls normally on the film air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is `Zxx10^(-7)m`. Then find `Z`.

Options

A. `10^(-7)`m
B. `2xx10^(-7) m`
C. `3xx10^(-7)m`
D. `5xx10^(-7)m`

Explanation

To solve the problem, we need to find the minimum change in thickness of a thin oil film that causes it to appear bright when viewed normally after initially appearing dark.

### Step-by-Step Solution:

1. **Understanding the Condition for Dark Appearance**:
- The film appears dark when the path difference for the reflected rays is an odd multiple of half the wavelength. This condition can be expressed as:
\[ \Delta x = (2n - 1) \frac{\lambda}{2} \]
- Here, \( n \) is an integer representing the order of interference, and \( \lambda \) is the wavelength of the light.

2. **Path Difference Calculation**:
- For a thin film of thickness \( T \) and refractive index \( \mu_1 \), the path difference for the rays reflected from the top and bottom surfaces of the film can be given by:
\[ \Delta x = 2 \mu_1 T \]
- Setting these equal for the dark condition:
\[ 2 \mu_1 T = (2n - 1) \frac{\lambda}{2} \]

3. **Solving for Thickness \( T \)**:
- Rearranging the equation gives:
\[ T = \frac{(2n - 1) \lambda}{4 \mu_1} \]

4. **Condition for Bright Appearance**:
- For the film to appear bright, the path difference must be an integer multiple of the wavelength:
\[ \Delta x = n \lambda \]
- Setting this for the bright condition:
\[ 2 \mu_1 T' = n \lambda \]
- Rearranging gives:
\[ T' = \frac{n \lambda}{2 \mu_1} \]

5. **Finding Change in Thickness**:
- The change in thickness \( \Delta T \) required for the film to change from dark to bright is:
\[ \Delta T = T' - T \]
- Substituting the expressions for \( T' \) and \( T \):
\[ \Delta T = \frac{n \lambda}{2 \mu_1} - \frac{(2n - 1) \lambda}{4 \mu_1} \]
- Finding a common denominator and simplifying:
\[ \Delta T = \frac{2n \lambda - (2n - 1) \lambda}{4 \mu_1} = \frac{(2n - 2n + 1) \lambda}{4 \mu_1} = \frac{\lambda}{4 \mu_1} \]

6. **Substituting Values**:
- Given \( \lambda = 9.6 \times 10^{-7} \, m \) and \( \mu_1 = 1.2 \):
\[ \Delta T = \frac{9.6 \times 10^{-7}}{4 \times 1.2} \]
- Calculating:
\[ \Delta T = \frac{9.6 \times 10^{-7}}{4.8} = 2 \times 10^{-7} \, m \]

7. **Final Result**:
- The minimum change in thickness \( \Delta T \) is \( 2 \times 10^{-7} \, m \).
- Therefore, \( Z = 2 \).