To solve the problem, we need to determine the wavelengths of light that will not be seen in the reflected light from an oil film of thickness \(10^{-4} \, \text{cm}\) and refractive index \(1.4\). This situation involves the phenomenon of thin film interference.
### Step-by-Step Solution:
1. **Convert Thickness to Meters**:
The thickness \(T\) of the oil film is given as \(10^{-4} \, \text{cm}\). We need to convert this to meters:
\[
T = 10^{-4} \, \text{cm} = 10^{-4} \times 10^{-2} \, \text{m} = 10^{-6} \, \text{m}
\]
2. **Determine the Path Difference**:
For a thin film of thickness \(T\) and refractive index \(\mu\), the effective path difference for light reflecting off the top and bottom surfaces of the film is given by:
\[
\text{Path Difference} = 2 \mu T
\]
Substituting the values:
\[
\text{Path Difference} = 2 \times 1.4 \times 10^{-6} \, \text{m} = 2.8 \times 10^{-6} \, \text{m}
\]
3. **Condition for Wavelength Not Seen**:
The wavelengths that will not be seen in the reflected light correspond to destructive interference. This occurs when the path difference is equal to an odd multiple of half the wavelength:
\[
\text{Path Difference} = (n + \frac{1}{2}) \lambda
\]
Rearranging gives:
\[
\lambda = \frac{2.8 \times 10^{-6}}{n + \frac{1}{2}}
\]
4. **Calculate Wavelengths for Different Values of \(n\)**:
We will calculate \(\lambda\) for several integer values of \(n\):
- For \(n = 0\):
\[
\lambda = \frac{2.8 \times 10^{-6}}{0 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{0.5} = 5.6 \times 10^{-6} \, \text{m} = 5600 \, \text{Å}
\]
- For \(n = 1\):
\[
\lambda = \frac{2.8 \times 10^{-6}}{1 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{1.5} = 1.86667 \times 10^{-6} \, \text{m} \approx 1866.67 \, \text{Å}
\]
- For \(n = 2\):
\[
\lambda = \frac{2.8 \times 10^{-6}}{2 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{2.5} = 1.12 \times 10^{-6} \, \text{m} = 1120 \, \text{Å}
\]
- For \(n = 3\):
\[
\lambda = \frac{2.8 \times 10^{-6}}{3 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{3.5} = 0.8 \times 10^{-6} \, \text{m} = 800 \, \text{Å}
\]
5. **Conclusion**:
The wavelengths that will not be seen in the reflected light due to destructive interference are approximately \(5600 \, \text{Å}\), \(1866.67 \, \text{Å}\), \(1120 \, \text{Å}\), and \(800 \, \text{Å}\). Since the question asks for wavelengths that will not be seen, all calculated wavelengths correspond to conditions of destructive interference.
### Final Answer:
The wavelengths that will not be seen in the reflected light are \(5600 \, \text{Å}\), \(1866.67 \, \text{Å}\), \(1120 \, \text{Å}\), and \(800 \, \text{Å}\).
