A soap film of thickness is surrounded by air. It is illuminated at near normal incidence by monochromatic light which has wavelength `lambda` in the film. What will be the film thickness that will produce maximum brightness of the reflected light?

Options

A. `(1 // 4)lambda`
B. `(1 // 2)lambda`
C. `lambda`
D. `2lambda`

Answer

To find the thickness of a soap film that produces maximum brightness of the reflected light, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Interference Condition**:
- When light reflects off a soap film, it undergoes interference. For constructive interference (maximum brightness), the path difference between the light reflected from the top and bottom surfaces of the film must satisfy certain conditions.

2. **Path Difference Calculation**:
- The path difference for light reflecting off a thin film is given by:
\[ \text{Path Difference} = 2nt \cos(r) \]
- Here, \( n \) is the refractive index of the film, \( t \) is the thickness of the film, and \( r \) is the angle of refraction. Since the light is incident normally, \( r = 0 \) and \( \cos(0) = 1 \). Therefore, the path difference simplifies to:
\[ \text{Path Difference} = 2nt \]

3. **Condition for Maximum Brightness**:
- For maximum brightness (constructive interference), the condition is:
\[ 2nt = (m + \frac{1}{2})\lambda \]
- Where \( m \) is an integer (0, 1, 2, ...). However, for the first maximum (brightest), we can take \( m = 0 \):
\[ 2nt = \frac{1}{2}\lambda \]

4. **Solving for Film Thickness**:
- Rearranging the equation to find \( t \):
\[ t = \frac{\lambda}{4n} \]

5. **Conclusion**:
- Thus, the thickness of the soap film that will produce maximum brightness of the reflected light is:
\[ t = \frac{\lambda}{4n} \]