A thin film of soap solution `(mu_(s)=1.5)`. When visible light is incident almost normal to the plate, two wavelength 420 nm and 630 nm are strongly reflected. What is the minimum thickness (in nm) of the soap film?

Answer

To solve the problem of finding the minimum thickness of a soap film that reflects two wavelengths of light (420 nm and 630 nm), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reflection Condition**:
When light reflects off a soap film, a phase change of π (or half a wavelength) occurs at the boundary where light moves from a less dense medium (air) to a denser medium (soap). This phase change affects the interference pattern.

2. **Condition for Constructive Interference**:
The condition for constructive interference (maximum intensity) in a thin film is given by:
\[ 2 \mu t = n \lambda \]
where:
- \( \mu \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( n \) is the order of interference (an integer),
- \( \lambda \) is the wavelength of light in the medium.

3. **Setting Up the Equations**:
For the two wavelengths (420 nm and 630 nm), we can set up the following equations based on the constructive interference condition:
\[ 2 \mu t = n_1 \cdot 420 \quad \text{(1)} \]
\[ 2 \mu t = n_2 \cdot 630 \quad \text{(2)} \]
Here, \( n_1 \) and \( n_2 \) are the orders of interference for the respective wavelengths.

4. **Equating the Two Expressions**:
Since both expressions equal \( 2 \mu t \), we can set them equal to each other:
\[ n_1 \cdot 420 = n_2 \cdot 630 \]

5. **Solving for the Ratio of Orders**:
Rearranging gives us:
\[ \frac{n_1}{n_2} = \frac{630}{420} = \frac{3}{2} \]
This means that for every 3 orders of interference for the 420 nm wavelength, there are 2 orders for the 630 nm wavelength.

6. **Finding the Minimum Thickness**:
Let \( n_1 = 3k \) and \( n_2 = 2k \) for some integer \( k \). Substituting back into equation (1):
\[ 2 \mu t = 3k \cdot 420 \]
and from equation (2):
\[ 2 \mu t = 2k \cdot 630 \]
Setting these equal gives:
\[ 3k \cdot 420 = 2k \cdot 630 \]
Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ 3 \cdot 420 = 2 \cdot 630 \]
This confirms our ratio is correct.

7. **Substituting for Thickness**:
Now we can find \( t \) using either equation. Using \( n_1 = 3 \):
\[ 2 \cdot 1.5 \cdot t = 3 \cdot 420 \]
\[ 3t = 1260 \implies t = \frac{1260}{3} = 420 \, \text{nm} \]

### Final Answer:
The minimum thickness of the soap film is **420 nm**.