A double slit of separation 0.1 mm is illuminated by white light. A coloured 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 step by step, 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 in a double-slit interference pattern.

### Step-by-Step Solution:

1. **Understand the Condition for Destructive Interference**:
Destructive interference occurs when the path difference between the light waves from the two slits is an odd multiple of half the wavelength. The condition for destructive interference is given by:
\[ \Delta y = \frac{(2n - 1) \lambda D}{d} \]
where:
- \( \Delta y \) is the distance from the central maximum to the point of interest (2 mm in this case),
- \( D \) is the distance from the slits to the screen (100 cm or 1 m),
- \( d \) is the distance between the slits (0.1 mm or \( 0.1 \times 10^{-3} \) m),
- \( n \) is an integer (1, 2, 3,...).

2. **Convert Units**:
Convert all measurements to meters for consistency:
- \( \Delta y = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \)
- \( d = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} \)
- \( D = 100 \text{ cm} = 1 \text{ m} \)

3. **Substitute Values into the Destructive Interference Condition**:
Rearranging the equation for \( \lambda \):
\[ \lambda = \frac{(2n - 1) \Delta y d}{D} \]
Substitute the known values:
\[ \lambda = \frac{(2n - 1) (2 \times 10^{-3}) (0.1 \times 10^{-3})}{1} \]
Simplifying this gives:
\[ \lambda = (2n - 1) \times 2 \times 10^{-7} \text{ m} \]

4. **Convert Wavelength to Angstroms**:
Since \( 1 \text{ m} = 10^{10} \text{ Å} \):
\[ \lambda = (2n - 1) \times 2 \times 10^{-7} \times 10^{10} \text{ Å} = (2n - 1) \times 2000 \text{ Å} \]

5. **Calculate Wavelengths for Different Values of n**:
- For \( n = 1 \):
\[ \lambda = (2 \times 1 - 1) \times 2000 = 2000 \text{ Å} \]
- For \( n = 2 \):
\[ \lambda = (2 \times 2 - 1) \times 2000 = 6000 \text{ Å} \]
- For \( n = 3 \):
\[ \lambda = (2 \times 3 - 1) \times 2000 = 10000 \text{ Å} \quad (\text{not in visible range}) \]

6. **Identify Wavelengths in the Visible Spectrum**:
The visible spectrum ranges from 4000 Å to 7000 Å. The wavelengths calculated that fall within this range are:
- \( 2000 \text{ Å} \) (not visible)
- \( 6000 \text{ Å} \) (visible)

7. **Conclusion**:
The wavelength that will be absent in the light transmitted through the pinhole is **6000 Å**.