E6. For the same conditions described in exercise 5, at what distance from the center of the screen will the fourth dark fringe appear?

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To determine the distance from the center of the screen where the **fourth dark fringe** appears, we need to use the formula for the position of dark fringes in a double-slit interference pattern. However, since the specific conditions from **Exercise 5** are not provided, I'll outline the general approach and formula you can use once you have the necessary parameters.

### General Formula for Dark Fringes

The position $$ y_m $$ of the $$ m^{th} $$ dark fringe in a double-slit interference pattern is given by:

$$
y_m = \left( m + \frac{1}{2} \right) \frac{\lambda D}{d}
$$

Where:
- $$ y_m $$ = distance from the center of the screen to the $$ m^{th} $$ dark fringe
- $$ m $$ = order of the dark fringe (for the fourth dark fringe, $$ m = 3 $$ since $$ m $$ starts from 0)
- $$ \lambda $$ = wavelength of the light used
- $$ D $$ = distance from the double-slit to the screen
- $$ d $$ = separation between the two slits

### Steps to Calculate

1. **Identify the Parameters:**
   - **Wavelength ($$ \lambda $$)**: This should be given in Exercise 5.
   - **Distance to Screen ($$ D $$)**: Also provided in Exercise 5.
   - **Slit Separation ($$ d $$)**: Provided in Exercise 5.

2. **Determine the Order ($$ m $$):**
   - For the fourth dark fringe, $$ m = 3 $$ (since $$ m = 0 $$ corresponds to the first dark fringe).

3. **Plug the Values into the Formula:**
   $$
   y_3 = \left( 3 + \frac{1}{2} \right) \frac{\lambda D}{d} = \frac{7}{2} \cdot \frac{\lambda D}{d}
   $$

4. **Calculate $$ y_3 $$:**
   - Substitute the known values of $$ \lambda $$, $$ D $$, and $$ d $$ into the equation to find the distance from the center to the fourth dark fringe.

### Example Calculation

*Assuming* the following values (replace these with the actual values from Exercise 5):
- Wavelength, $$ \lambda = 600 $$ nm $$ = 600 \times 10^{-9} $$ m
- Distance to screen, $$ D = 2 $$ m
- Slit separation, $$ d = 0.01 $$ m

$$
y_3 = \frac{7}{2} \cdot \frac{600 \times 10^{-9} \, \text{m} \times 2 \, \text{m}}{0.01 \, \text{m}} = \frac{7}{2} \cdot \frac{1.2 \times 10^{-6} \, \text{m}^2}{0.01 \, \text{m}} = \frac{7}{2} \times 1.2 \times 10^{-4} \, \text{m} = 4.2 \times 10^{-4} \, \text{m} = 0.42 \, \text{mm}
$$

So, in this example, the fourth dark fringe would appear **0.42 mm** from the center of the screen.

### Conclusion

To find the exact distance for your specific problem:

1. Refer back to **Exercise 5** to obtain the values for $$ \lambda $$, $$ D $$, and $$ d $$.
2. Use the formula provided above with $$ m = 3 $$ for the fourth dark fringe.
3. Perform the calculation to find $$ y_3 $$.

If you provide the specific values from Exercise 5, I can help you carry out the calculation more precisely!
To find the distance from the center of the screen where the fourth dark fringe appears, use the formula: $$ y_m = \left( m + \frac{1}{2} \right) \frac{\lambda D}{d} $$ Where: - $$ y_m $$ = distance to the $$ m^{th} $$ dark fringe - $$ m = 3 $$ for the fourth fringe - $$ \lambda $$ = wavelength - $$ D $$ = distance from slits to screen - $$ d $$ = slit separation Plug in the known values and calculate $$ y_3 $$ to find the exact distance.

E6. For the same conditions described in exercise 5, at what distance from the center of the screen will the fourth dark fringe appear?

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