A double slit of separation 0.1 mm is illuminated by white light. A colored interference pattern is formed on a screen 100 cm away. If a pin hole is located in this screen at a distance of 2 mm from the central fringe, the wavelength in the visible spectrum (4000 Å to 7000Å) which will be absent in the light transmitted through the pin hole is (are)

Options

A. 4000 Å
B. 5000Å
C. 6000Å
D. 7000Å

Explanation

To solve the problem, we need to determine the wavelengths of light that will be absent in the light transmitted through the pinhole located at a distance of 2 mm from the central fringe of the interference pattern formed by the double slit.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Slit separation (d) = 0.1 mm = \(0.1 \times 10^{-3}\) m
- Distance from slits to screen (D) = 100 cm = 1 m
- Distance from central fringe to pinhole (y) = 2 mm = \(2 \times 10^{-3}\) m

2. **Understand the Condition for Minima:**
The condition for minima in a double-slit interference pattern is given by:
\[ \Delta x = \frac{y \cdot d}{D} \]
where \(\Delta x\) is the path difference, and for minima, it is given by:
\[ \Delta x = \left(2n - 1\right) \frac{\lambda}{2} \]
where \(n\) is the order of the minimum.

3. **Set Up the Equation:**
Equating the two expressions for path difference:
\[ \frac{y \cdot d}{D} = \left(2n - 1\right) \frac{\lambda}{2} \]

4. **Rearranging for Wavelength (\(\lambda\)):**
Rearranging the equation gives:
\[ \lambda = \frac{2y \cdot d}{D(2n - 1)} \]

5. **Substituting Known Values:**
Substitute \(y\), \(d\), and \(D\) into the equation:
\[ \lambda = \frac{2 \cdot (2 \times 10^{-3}) \cdot (0.1 \times 10^{-3})}{1 \cdot (2n - 1)} \]
Simplifying this gives:
\[ \lambda = \frac{4 \times 10^{-6} \cdot 10^{-4}}{(2n - 1)} = \frac{4 \times 10^{-10}}{(2n - 1)} \text{ m} \]

6. **Convert to Angstroms:**
Since \(1 \text{ m} = 10^{10} \text{ Å}\), we convert \(\lambda\) to Angstroms:
\[ \lambda = \frac{4 \times 10^{-10} \times 10^{10}}{(2n - 1)} = \frac{4000}{(2n - 1)} \text{ Å} \]

7. **Finding Wavelengths in the Visible Spectrum:**
We need to find values of \(n\) such that \(\lambda\) falls within the visible spectrum range (4000 Å to 7000 Å):
- For \(n = 1\):
\[ \lambda = \frac{4000}{1} = 4000 \text{ Å} \]
- For \(n = 2\):
\[ \lambda = \frac{4000}{3} \approx 1333.33 \text{ Å} \quad (\text{Not in visible range}) \]
- For \(n = 3\):
\[ \lambda = \frac{4000}{5} = 800 \text{ Å} \quad (\text{Not in visible range}) \]
- For \(n = 4\):
\[ \lambda = \frac{4000}{7} \approx 571.43 \text{ Å} \quad (\text{Not in visible range}) \]
- For \(n = 5\):
\[ \lambda = \frac{4000}{9} \approx 444.44 \text{ Å} \quad (\text{Not in visible range}) \]
- For \(n = 6\):
\[ \lambda = \frac{4000}{11} \approx 363.64 \text{ Å} \quad (\text{Not in visible range}) \]

8. **Conclusion:**
The only wavelength that will be absent in the light transmitted through the pinhole is:
\[ \lambda = 4000 \text{ Å} \]