In YDSE for wavelength `lambda = 589nm`, the interference fringe have angular separation of `3050 xx 10^(-3)` rad. For what wavelength would the angular separation be 10.0% greater?

Explanation

To solve the problem, we need to find the new wavelength that results in an angular separation that is 10% greater than the given angular separation for the original wavelength of 589 nm.

### Step-by-Step Solution:

1. **Understanding Angular Separation**:
The angular separation (Δθ) between the fringes in a Young's double slit experiment is given by the formula:
\[
\Delta \theta = \frac{\lambda}{d}
...