In Young's double slit experiement when wavelength used is `6000Å` and the screen is `40cm` from the slits, the fringes are `0.012cm` wide. What is the distance between the slits?

Options

A. (a) `0.024cm`
B. (b) `2.4cm`
C. (c) `0.24cm`
D. (d) `0.2cm`

Explanation

To solve the problem, we will use the formula for fringe width in Young's double slit experiment. The formula is given by:

\[ \beta = \frac{\lambda D}{d} \]

Where:
- \(\beta\) = fringe width
- \(\lambda\) = wavelength of light
- \(D\) = distance from the slits to the screen
- \(d\) = distance between the slits

We need to rearrange this formula to solve for \(d\):

\[ d = \frac{\lambda D}{\beta} \]

Now, let's plug in the values provided in the question.

1. **Convert the wavelength from angstroms to centimeters**:
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6000 \times 10^{-8} \, \text{cm} = 6 \times 10^{-6} \, \text{cm} \]

2. **Distance from the slits to the screen**:
\[ D = 40 \, \text{cm} \]

3. **Fringe width**:
\[ \beta = 0.012 \, \text{cm} \]

4. **Substituting the values into the formula**:
\[ d = \frac{(6 \times 10^{-6} \, \text{cm}) \times (40 \, \text{cm})}{0.012 \, \text{cm}} \]

5. **Calculating \(d\)**:
\[ d = \frac{240 \times 10^{-6}}{0.012} \]
\[ d = \frac{240 \times 10^{-6}}{0.012} = 20 \times 10^{-3} \, \text{cm} = 0.02 \, \text{cm} \]

Thus, the distance between the slits \(d\) is \(0.02 \, \text{cm}\).

### Final Answer:
The distance between the slits is \(0.02 \, \text{cm}\).