To solve the problem of finding the missing wavelength in a Young's double slit experiment with white light, we can follow these steps:
### Step-by-step Solution:
1. **Understanding the Setup**:
- We have a double-slit experiment where white light (wavelength range 400 nm to 700 nm) is incident normally on the slits.
- The distance between the slits \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).
- The distance from the slits to the screen \( D = 0.8 \, \text{m} \).
2. **Path Difference**:
- The path difference \( \Delta x \) for light coming from the two slits to a point on the screen can be expressed as:
\[
\Delta x = d \sin \theta
\]
- For small angles, \( \sin \theta \approx \tan \theta \), where \( \tan \theta = \frac{y}{D} \) (with \( y \) being the vertical distance from the central maximum to the point on the screen).
3. **Expression for Path Difference**:
- Thus, we can write:
\[
\Delta x = d \frac{y}{D}
\]
4. **Condition for Dark Fringes**:
- The condition for dark fringes (missing wavelengths) is given by:
\[
\Delta x = \left( n + \frac{1}{2} \right) \lambda
\]
- Here, \( n \) is an integer (0, 1, 2, ...), and \( \lambda \) is the wavelength.
5. **Equating the Two Expressions**:
- Setting the two expressions for path difference equal gives:
\[
d \frac{y}{D} = \left( n + \frac{1}{2} \right) \lambda
\]
6. **Finding Wavelength**:
- Rearranging gives:
\[
\lambda = \frac{d y}{D \left( n + \frac{1}{2} \right)}
\]
7. **Calculating Wavelength for Missing Values**:
- Since we are looking for the missing wavelengths, we need to find \( \lambda \) for different values of \( n \).
- Substitute \( d = 1 \times 10^{-3} \, \text{m} \) and \( D = 0.8 \, \text{m} \):
\[
\lambda = \frac{(1 \times 10^{-3}) y}{0.8 \left( n + \frac{1}{2} \right)}
\]
8. **Finding Specific Values**:
- For \( n = 0 \):
\[
\lambda = \frac{(1 \times 10^{-3}) y}{0.8 \times 0.5} = \frac{(1 \times 10^{-3}) y}{0.4} = 2.5 \times 10^{-3} y
\]
- For \( n = 1 \):
\[
\lambda = \frac{(1 \times 10^{-3}) y}{0.8 \times 1.5} = \frac{(1 \times 10^{-3}) y}{1.2} \approx 8.33 \times 10^{-4} y
\]
- Continue this for \( n = 2, 3, \ldots \) until the calculated wavelengths fall within the range of 400 nm to 700 nm.
9. **Final Calculation**:
- After calculating for different \( n \), we find that the first missing wavelength in the range is approximately \( 416.67 \, \text{nm} \) for \( n = 1 \).
### Conclusion:
The missing wavelength is approximately **416.67 nm**, which falls within the visible spectrum of white light.
