A double-slit experiment, arrangement produces interference fringes for sodium light `(lambda = 589 nm)` that have an angular separation of `3.50 xx 0^(-3)` rad . For what wavelength would the angular separation to `10%` greater?

Options

A. 527 nm
B. 648 nm
C. 722 nm
D. 449 nm

Explanation

To solve the problem, we need to find the new wavelength for which the angular separation is 10% greater than the original angular separation produced by sodium light.

### Step-by-Step Solution:

1. **Identify the given values:**
- Original wavelength, \( \lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \)
- Original angular separation, \( \omega = 3.50 \times 10^{-3} \, \text{rad} \)

2. **Determine the new angular separation:**
- The new angular separation, \( \omega' \), is 10% greater than the original:
\[ \omega' = \omega + 0.1 \omega = 1.1 \omega \]
- Thus,
\[ \omega' = 1.1 \times (3.50 \times 10^{-3}) = 3.85 \times 10^{-3} \, \text{rad} \]

3. **Use the relationship between angular separation and wavelength:**
- The angular separation in a double-slit experiment is given by:
\[ \omega = \frac{\lambda}{d} \]
- Therefore, for the new wavelength \( \lambda' \):
\[ \omega' = \frac{\lambda'}{d} \]

4. **Set up the ratio of angular separations:**
- From the above equations, we can set up the ratio:
\[ \frac{\omega'}{\omega} = \frac{\lambda'}{\lambda} \]
- Substituting \( \omega' = 1.1 \omega \):
\[ \frac{1.1 \omega}{\omega} = \frac{\lambda'}{\lambda} \]
- This simplifies to:
\[ 1.1 = \frac{\lambda'}{\lambda} \]

5. **Solve for the new wavelength \( \lambda' \):**
- Rearranging gives:
\[ \lambda' = 1.1 \lambda \]
- Substituting the value of \( \lambda \):
\[ \lambda' = 1.1 \times 589 \, \text{nm} = 647.9 \, \text{nm} \]

6. **Round the answer:**
- Rounding \( 647.9 \, \text{nm} \) gives approximately:
\[ \lambda' \approx 648 \, \text{nm} \]

### Final Answer:
The new wavelength for which the angular separation is 10% greater is approximately \( 648 \, \text{nm} \).