In Young's double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular `d_(0)` between the slits. If the angular resolution of the eye is 1 °/60, the value of `d_(0)` is?

Options

A. 1 mm
B. 3 mm
C. 2 mm
D. 4 mm

Answer

To solve the problem, we need to find the value of \( d_0 \) (the separation between the slits) at which the interference pattern disappears for a person looking at it. We will use the concept of angular resolution and the fringe width in Young's double slit experiment.

### Step-by-step Solution:

1. **Identify Given Values**:
- Distance between slits and screen, \( D = 1 \, \text{m} \)
- Wavelength of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Angular resolution of the eye, \( \theta = \frac{1}{60} \, \text{degrees} \)

2. **Convert Angular Resolution to Radians**:
- To convert degrees to radians, use the conversion factor \( \frac{\pi \, \text{radians}}{180 \, \text{degrees}} \).
\[ \theta = \frac{1}{60} \times \frac{\pi}{180} = \frac{\pi}{10800} \, \text{radians} \]

3. **Understand the Condition for Disappearance of the Pattern**:
- The interference pattern disappears when the angular width of the fringe becomes equal to the angular resolution of the eye.
- The angular width \( \theta \) can be expressed in terms of fringe width \( \beta \) and distance \( D \):
\[ \theta = \frac{\beta}{D} \]
where fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \]
Here, \( d \) is the separation between the slits.

4. **Substituting for Fringe Width**:
- Substitute \( \beta \) in the angular width equation:
\[ \theta = \frac{\lambda D}{d D} = \frac{\lambda}{d} \]

5. **Setting Angular Resolution Equal to Angular Width**:
- Set the angular resolution equal to the angular width:
\[ \frac{\pi}{10800} = \frac{600 \times 10^{-9}}{d_0} \]

6. **Solving for \( d_0 \)**:
- Rearranging the equation gives:
\[ d_0 = \frac{600 \times 10^{-9} \times 10800}{\pi} \]
- Calculating \( d_0 \):
\[ d_0 \approx \frac{600 \times 10^{-9} \times 10800}{3.14159} \approx 2.06 \times 10^{-3} \, \text{m} = 2.06 \, \text{mm} \]

### Final Answer:
The value of \( d_0 \) is approximately \( 2.06 \, \text{mm} \).

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