$$ d \sin \theta = m \lambda $$

For two wavelengths, $$\lambda_1 = 600\,\text{nm}$$ and $$\lambda_2 = 480\,\text{nm}$$, the bright fringes coincide when the conditions

$$ m \lambda_2 = n \lambda_1 $$

are simultaneously satisfied for integers $$m$$ and $$n$$, representing the bright fringe order for each wavelength.

Rearranging the equation gives:

$$ \frac{m}{n} = \frac{\lambda_1}{\lambda_2} = \frac{600}{480} = \frac{5}{4} $$

Thus, the smallest positive integers that satisfy this ratio are:

$$ m = 5 \quad \text{and} \quad n = 4. $$

This means the first coincidence of the bright fringes occurs at the 5th bright fringe for the 480 nm light.