White light falls normally on a film of soap water whose thickness is `5xx10^(-5)cm` and refractive index is 1.40. The wavelengths in the visible region that are reflected the most strongly are :

Options

A. 5000 Å and 4000 Å
B. 5400 Å and 4000 Å
C. 6000 Å and 5000 Å
D. 4500 Å only

Explanation

To solve the problem, we need to find the wavelengths of light that are reflected most strongly from a soap film of given thickness and refractive index. This involves understanding the interference of light waves reflecting off the top and bottom surfaces of the soap film.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Thickness of the soap film, \( t = 5 \times 10^{-5} \) cm = \( 5 \times 10^{-7} \) m.
- Refractive index of the soap film, \( n = 1.40 \).

2. **Convert the Thickness to Meters**:
- Since the thickness is given in centimeters, we convert it to meters for standard SI units:
\[ t = 5 \times 10^{-5} \text{ cm} = 5 \times 10^{-7} \text{ m} \]

3. **Determine the Condition for Constructive Interference**:
- For constructive interference in a thin film, the condition is given by:
\[ 2nt = (m + \frac{1}{2})\lambda \]
where \( m \) is an integer (0, 1, 2, ...), and \( \lambda \) is the wavelength of light in vacuum.

4. **Calculate the Effective Wavelength in the Film**:
- The wavelength of light in the film is given by:
\[ \lambda_{film} = \frac{\lambda}{n} \]

5. **Substituting into the Condition**:
- Rearranging the equation for constructive interference:
\[ \lambda = \frac{2nt}{m + \frac{1}{2}} \]

6. **Substituting the Known Values**:
- Substitute \( n = 1.40 \) and \( t = 5 \times 10^{-7} \) m into the equation:
\[ \lambda = \frac{2 \times 1.40 \times 5 \times 10^{-7}}{m + \frac{1}{2}} \]
\[ \lambda = \frac{1.4 \times 10^{-6}}{m + \frac{1}{2}} \]

7. **Finding Wavelengths in the Visible Range**:
- The visible range of wavelengths is approximately \( 400 \) nm to \( 700 \) nm (or \( 4 \times 10^{-7} \) m to \( 7 \times 10^{-7} \) m).
- We need to find integer values of \( m \) such that \( \lambda \) falls within this range.

8. **Calculate for Different Values of \( m \)**:
- For \( m = 0 \):
\[ \lambda = \frac{1.4 \times 10^{-6}}{0.5} = 2.8 \times 10^{-6} \text{ m} \quad (\text{not visible}) \]
- For \( m = 1 \):
\[ \lambda = \frac{1.4 \times 10^{-6}}{1.5} \approx 9.33 \times 10^{-7} \text{ m} \quad (\text{not visible}) \]
- For \( m = 2 \):
\[ \lambda = \frac{1.4 \times 10^{-6}}{2.5} = 5.6 \times 10^{-7} \text{ m} = 560 \text{ nm} \quad (\text{visible}) \]
- For \( m = 3 \):
\[ \lambda = \frac{1.4 \times 10^{-6}}{3.5} \approx 4.0 \times 10^{-7} \text{ m} = 400 \text{ nm} \quad (\text{visible}) \]
- For \( m = 4 \):
\[ \lambda = \frac{1.4 \times 10^{-6}}{4.5} \approx 3.11 \times 10^{-7} \text{ m} \quad (\text{not visible}) \]

9. **Conclusion**:
- The wavelengths in the visible region that are reflected most strongly are approximately \( 400 \) nm and \( 560 \) nm.