When light of wavelength `4000overset(o)(A)` in vacuum travels through the same thickness in air and vacuum the difference in the number of waves is one. Find the thickness (`mu_(air)` =1.0008).

Options

A. 0.5mm
B. 1mm
C. 18cm
D. 24cm

Explanation

To solve the problem, we need to find the thickness \( T \) of the medium (air) through which light of wavelength \( 4000 \, \text{Å} \) travels, given that the difference in the number of waves traveling through the same thickness in air and vacuum is one.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The wavelength of light in vacuum \( \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \).
- The refractive index of air \( \mu_{\text{air}} = 1.0008 \).
- The difference in the number of waves \( n_{\text{vacuum}} - n_{\text{air}} = 1 \).

2. **Calculate the Number of Waves**:
- The number of waves in a medium is given by the formula:
\[ n = \frac{d}{\lambda} \]
- For air, the effective wavelength \( \lambda_{\text{air}} \) can be calculated as:
\[ \lambda_{\text{air}} = \frac{\lambda}{\mu_{\text{air}}} = \frac{4000 \times 10^{-10}}{1.0008} \]

3. **Expressing the Number of Waves**:
- The number of waves in air:
\[ n_{\text{air}} = \frac{T}{\lambda_{\text{air}}} \]
- The number of waves in vacuum:
\[ n_{\text{vacuum}} = \frac{T}{\lambda} \]

4. **Setting up the Equation**:
- From the problem statement, we have:
\[ n_{\text{vacuum}} - n_{\text{air}} = 1 \]
- Substituting the expressions for \( n_{\text{vacuum}} \) and \( n_{\text{air}} \):
\[ \frac{T}{\lambda} - \frac{T}{\lambda_{\text{air}}} = 1 \]

5. **Substituting for \( \lambda_{\text{air}} \)**:
- Substitute \( \lambda_{\text{air}} = \frac{\lambda}{\mu_{\text{air}}} \):
\[ \frac{T}{\lambda} - \frac{T \mu_{\text{air}}}{\lambda} = 1 \]
- Simplifying gives:
\[ T \left( \frac{1 - \mu_{\text{air}}}{\lambda} \right) = 1 \]

6. **Solving for Thickness \( T \)**:
- Rearranging gives:
\[ T = \frac{\lambda}{1 - \mu_{\text{air}}} \]
- Substitute \( \mu_{\text{air}} = 1.0008 \):
\[ T = \frac{4000 \times 10^{-10}}{1 - 1.0008} = \frac{4000 \times 10^{-10}}{-0.0008} \]
- Calculating \( T \):
\[ T = \frac{4000 \times 10^{-10}}{-0.0008} = -5 \times 10^{-6} \, \text{m} = 0.5 \, \text{mm} \]

### Final Answer:
The thickness \( T \) is \( 0.5 \, \text{mm} \).