A parallel beam of light of all wavelength greater than 3000Å falls on a double slit in a Young's double slit experiment. It is observed that the wavelength 3600Å and 6000Å are absent at a distance of 31.5 mm from the position of the centre maximum and the orders of the interference at this point for the two wavelength differ by 7. If the distance between the slit and the screen in 1m, the separation between the two slits is

Options

A. 0.08 mm
B. 0.13 mm
C. 0.2 mm
D. 0.1 mm

Answer

To solve the problem, we need to find the separation between the two slits (denoted as \(d\)) in a Young's double slit experiment. We are given the following information:

1. The wavelengths of light are \(3600 \, \text{Å}\) and \(6000 \, \text{Å}\).
2. The distance from the center maximum to the point where these wavelengths are absent is \(y = 31.5 \, \text{mm} = 31.5 \times 10^{-3} \, \text{m}\).
3. The distance from the slits to the screen is \(D = 1 \, \text{m}\).
4. The difference in the orders of interference for the two wavelengths is \(n_2 - n_1 = 7\).

### Step-by-Step Solution:

**Step 1: Convert the wavelengths to meters.**
- Wavelengths given are in angstroms. Convert them to meters:
\[ \lambda_1 = 3600 \, \text{Å} = 3600 \times 10^{-10} \, \text{m} = 3.6 \times 10^{-7} \, \text{m} \]
\[ \lambda_2 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6.0 \times 10^{-7} \, \text{m} \]

**Step 2: Use the formula for fringe width.**
- The fringe width (\(\beta\)) is given by the formula:
\[ \beta = \frac{n \lambda D}{d} \]
where \(n\) is the order of the fringe.

**Step 3: Set up equations for both wavelengths.**
- For wavelength \(3600 \, \text{Å}\):
\[ y = \frac{n_1 \lambda_1 D}{d} \implies n_1 = \frac{y d}{\lambda_1 D} \]

- For wavelength \(6000 \, \text{Å}\):
\[ y = \frac{n_2 \lambda_2 D}{d} \implies n_2 = \frac{y d}{\lambda_2 D} \]

**Step 4: Relate the two orders of interference.**
- Since \(n_2 - n_1 = 7\), we can write:
\[ \frac{y d}{\lambda_2 D} - \frac{y d}{\lambda_1 D} = 7 \]
Factoring out \(\frac{y d}{D}\):
\[ \frac{y d}{D} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) = 7 \]

**Step 5: Substitute known values.**
- Substitute \(y = 31.5 \times 10^{-3} \, \text{m}\), \(D = 1 \, \text{m}\), \(\lambda_1 = 3.6 \times 10^{-7} \, \text{m}\), and \(\lambda_2 = 6.0 \times 10^{-7} \, \text{m}\):
\[ \frac{31.5 \times 10^{-3} \cdot d}{1} \left( \frac{1}{6.0 \times 10^{-7}} - \frac{1}{3.6 \times 10^{-7}} \right) = 7 \]

**Step 6: Calculate the difference in wavelengths.**
- Calculate the difference:
\[ \frac{1}{6.0 \times 10^{-7}} - \frac{1}{3.6 \times 10^{-7}} = \frac{3.6 - 6.0}{21.6 \times 10^{-13}} = \frac{-2.4}{21.6 \times 10^{-13}} = -1.111 \times 10^{13} \]

**Step 7: Solve for \(d\).**
- Rearranging gives:
\[ 31.5 \times 10^{-3} \cdot d \cdot (-1.111 \times 10^{13}) = 7 \]
\[ d = \frac{7}{31.5 \times 10^{-3} \cdot (-1.111 \times 10^{13})} \]

**Step 8: Calculate \(d\).**
- Calculate \(d\):
\[ d \approx 0.222 \, \text{mm} \]

### Final Answer:
The separation between the two slits is approximately \(0.222 \, \text{mm}\).