Find the distance between third and first maxima. `(d = 0.5mm, D = 0.5m, lambda = 5890 A^0)`

Answer

To find the distance between the third and first maxima in a double-slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Distance between slits, \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Distance from slits to screen, \( D = 0.5 \, \text{m} \)
- Wavelength of light, \( \lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, \text{m} \)

2. **Formula for Position of Maxima**:
The position of the \( n \)-th maxima from the central maxima is given by:
\[ y_n = \frac{n \lambda D}{d} \]

3. **Calculate the Position of the Third Maxima**:
For the third maxima (\( n = 3 \)):
\[ y_3 = \frac{3 \lambda D}{d} \]

4. **Calculate the Position of the First Maxima**:
For the first maxima (\( n = 1 \)):
\[ y_1 = \frac{1 \lambda D}{d} \]

5. **Find the Distance Between the Third and First Maxima**:
The distance between the third and first maxima is:
\[ \Delta y = y_3 - y_1 = \frac{3 \lambda D}{d} - \frac{1 \lambda D}{d} \]
Simplifying this gives:
\[ \Delta y = \frac{(3 - 1) \lambda D}{d} = \frac{2 \lambda D}{d} \]

6. **Substitute the Values**:
Now substituting the known values:
\[ \Delta y = \frac{2 \times (5890 \times 10^{-10} \, \text{m}) \times (0.5 \, \text{m})}{0.5 \times 10^{-3} \, \text{m}} \]

7. **Calculate**:
\[ \Delta y = \frac{2 \times 5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}} \]
\[ = \frac{5890 \times 10^{-10} \times 2 \times 0.5}{0.5 \times 10^{-3}} = \frac{5890 \times 10^{-10} \times 1}{10^{-3}} = 5890 \times 10^{-7} \, \text{m} \]
\[ = 5.890 \times 10^{-4} \, \text{m} = 0.000589 \, \text{m} \]

8. **Final Result**:
The distance between the third and first maxima is:
\[ \Delta y = 1.178 \times 10^{-3} \, \text{m} = 1.178 \, \text{mm} \]