A parallel beam of white light is incident normally on a water film` 1.0 x 10^-4` cm thick. Find the wavelength in the visible range (400 nm-700 nm) which are strongly transmitted by the film Refractive index of water = 1.33.

Explanation

To solve the problem, we need to find the wavelengths of light that are strongly transmitted through a water film of thickness \(1.0 \times 10^{-4}\) cm and a refractive index of 1.33. We will use the formula for constructive interference in thin films.

### Step-by-Step Solution:

1. **Convert the thickness of the water film to meters:**
\[ T = 1.0 \times 10^{-4} \text{ cm} = 1.0 \times 10^{-6} \text{ m} \]

2. **Use the formula for constructive interference:**
The condition for constructive interference in a thin film is given by:
\[ 2 \mu T = n \lambda \]
where:
- \( \mu \) is the refractive index of the film (1.33 for water),
- \( T \) is the thickness of the film,
- \( n \) is the order of interference (an integer),
- \( \lambda \) is the wavelength of light in vacuum.

3. **Rearrange the formula to solve for \( \lambda \):**
\[ \lambda = \frac{2 \mu T}{n} \]

4. **Substitute the values into the equation:**
\[ \lambda = \frac{2 \times 1.33 \times 1.0 \times 10^{-6}}{n} \]
\[ \lambda = \frac{2.66 \times 10^{-6}}{n} \text{ m} \]
Converting to nanometers (1 m = \(10^9\) nm):
\[ \lambda = \frac{2.66 \times 10^{3}}{n} \text{ nm} \]

5. **Calculate \( \lambda \) for different values of \( n \):**
- For \( n = 1 \):
\[ \lambda_1 = \frac{2660}{1} = 2660 \text{ nm} \quad (\text{not in visible range}) \]
- For \( n = 2 \):
\[ \lambda_2 = \frac{2660}{2} = 1330 \text{ nm} \quad (\text{not in visible range}) \]
- For \( n = 3 \):
\[ \lambda_3 = \frac{2660}{3} \approx 886.67 \text{ nm} \quad (\text{not in visible range}) \]
- For \( n = 4 \):
\[ \lambda_4 = \frac{2660}{4} = 665 \text{ nm} \quad (\text{in visible range}) \]
- For \( n = 5 \):
\[ \lambda_5 = \frac{2660}{5} = 532 \text{ nm} \quad (\text{in visible range}) \]
- For \( n = 6 \):
\[ \lambda_6 = \frac{2660}{6} \approx 443.33 \text{ nm} \quad (\text{in visible range}) \]
- For \( n = 7 \):
\[ \lambda_7 = \frac{2660}{7} \approx 380 \text{ nm} \quad (\text{not in visible range}) \]

6. **Final wavelengths in the visible range:**
The wavelengths that are strongly transmitted by the film are:
- \( 665 \text{ nm} \)
- \( 532 \text{ nm} \)
- \( 443.33 \text{ nm} \)

### Summary of Results:
The wavelengths in the visible range that are strongly transmitted by the water film are:
- \( 665 \text{ nm} \)
- \( 532 \text{ nm} \)
- \( 443.33 \text{ nm} \)