A `4.0 xx 10^(2)` nm thick film of kerosene (n = 1.2) is floating on water. White light is normally incident on the film. What is the visible wavelength in air that has a maximum intensity after the light is reflected?
Note: The visible wavelength range is 380 nm to 750 nm.

Options

A. 380 nm
B. 480 nm
C. 430 nm
D. 530 nm

Answer

To find the visible wavelength in air that has a maximum intensity after being reflected from a thin film of kerosene, we can follow these steps:

### Step 1: Understand the Thin Film Interference
When light reflects off a thin film, interference occurs. For constructive interference (maximum intensity), the condition is given by:
\[ 2nt = (m + \frac{1}{2}) \lambda \]
for m = 0, 1, 2, ..., where:
- \( n \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( \lambda \) is the wavelength in the medium (kerosene in this case),
- \( m \) is the order of interference.

### Step 2: Calculate the Effective Wavelength in Kerosene
The wavelength of light in the medium can be calculated using the formula:
\[ \lambda_{medium} = \frac{\lambda_{air}}{n} \]
where \( n \) is the refractive index of kerosene (1.2).

### Step 3: Substitute into the Interference Condition
Substituting \( \lambda_{medium} \) into the interference condition gives:
\[ 2nt = (m + \frac{1}{2}) \frac{\lambda_{air}}{n} \]
Rearranging this, we find:
\[ \lambda_{air} = \frac{2nt^2}{m + \frac{1}{2}} \]

### Step 4: Substitute Known Values
Given:
- \( n = 1.2 \)
- \( t = 400 \, nm = 4.0 \times 10^2 \, nm \)

Substituting these values into the equation:
\[ \lambda_{air} = \frac{2 \times 1.2 \times 400}{m + \frac{1}{2}} \]

### Step 5: Calculate for Different Orders (m)
Now we will calculate \( \lambda_{air} \) for different integer values of \( m \) to find which wavelengths fall within the visible range (380 nm to 750 nm).

1. For \( m = 0 \):
\[ \lambda_{air} = \frac{2 \times 1.2 \times 400}{0 + \frac{1}{2}} = \frac{960}{0.5} = 1920 \, nm \] (not visible)

2. For \( m = 1 \):
\[ \lambda_{air} = \frac{2 \times 1.2 \times 400}{1 + \frac{1}{2}} = \frac{960}{1.5} = 640 \, nm \] (visible)

3. For \( m = 2 \):
\[ \lambda_{air} = \frac{2 \times 1.2 \times 400}{2 + \frac{1}{2}} = \frac{960}{2.5} = 384 \, nm \] (visible, but at the lower limit)

4. For \( m = 3 \):
\[ \lambda_{air} = \frac{2 \times 1.2 \times 400}{3 + \frac{1}{2}} = \frac{960}{3.5} \approx 274.29 \, nm \] (not visible)

### Step 6: Conclusion
The visible wavelengths that produce maximum intensity after reflection are:
- \( 640 \, nm \) (for \( m = 1 \))
- \( 384 \, nm \) (for \( m = 2 \))

However, since 384 nm is at the edge of the visible spectrum, the most prominent wavelength that has maximum intensity in the visible range is:
\[ \lambda_{air} = 640 \, nm \]