A thin oil film of refractive index 1.2 floats on the surface of water `(mu = 4/3)`. When a light of wavelength `lambda = 9.6xx 10^(-7)m` falls normally on the film from 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:

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Condition for Dark Appearance**:
When light reflects off a thin film, it can interfere constructively or destructively. For the film to appear dark, the condition for destructive interference must be satisfied. The path difference for destructive interference is given by:
\[ \Delta x = (2n - 1) \frac{\lambda}{2} \]
where \( n \) is an integer, and \( \lambda \) is the wavelength of the light.

2. **Path Difference Calculation**:
The path difference for light reflecting off the top and bottom surfaces of the oil film is given by:
\[ \Delta x = 2t \cdot n_1 \]
where \( t \) is the thickness of the film and \( n_1 \) is the refractive index of the oil film (1.2).

Since the light is reflecting off a medium of higher refractive index (water) from the oil, we have:
\[ 2t \cdot n_1 = (2n - 1) \frac{\lambda}{2} \]

3. **Condition for Bright Appearance**:
For the film to appear bright, the condition for constructive interference must be satisfied:
\[ \Delta x = n \lambda \]
Thus, for the new thickness \( t' \):
\[ 2t' \cdot n_1 = n \lambda \]

4. **Finding the Change in Thickness**:
We want to find the minimum change in thickness \( \Delta t = t' - t \). Rearranging the equations for \( t \) and \( t' \):
- From the dark condition:
\[ t = \frac{(2n - 1) \lambda}{4n_1} \]
- From the bright condition:
\[ t' = \frac{n \lambda}{2n_1} \]

Now substituting these into the change in thickness:
\[ \Delta t = t' - t = \left(\frac{n \lambda}{2n_1}\right) - \left(\frac{(2n - 1) \lambda}{4n_1}\right) \]
Simplifying:
\[ \Delta t = \frac{2n \lambda - (2n - 1) \lambda}{4n_1} = \frac{(2n - 2n + 1) \lambda}{4n_1} = \frac{\lambda}{4n_1} \]

5. **Substituting Values**:
Now, substituting \( \lambda = 9.6 \times 10^{-7} \, m \) and \( n_1 = 1.2 \):
\[ \Delta t = \frac{9.6 \times 10^{-7}}{4 \times 1.2} = \frac{9.6 \times 10^{-7}}{4.8} = 2 \times 10^{-7} \, m \]

### Final Answer:
The minimum change in thickness for which the film will appear bright in normally reflected light is:
\[ \Delta t = 2 \times 10^{-7} \, m \]