Young's double-slit experiment is performed with a 600 nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.5mm, from the central maximum . Determine the spacing of the slits.

Explanation

To solve the problem step by step, we can use the formula for the position of the minima in Young's double-slit experiment. The position of the nth minima is given by the formula:

\[ y_n = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} \]

Where:
- \(y_n\) is the position of the nth minima from the central maximum,
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