1. In Young’s double slit experiment, the position of the $n$-th bright fringe is $y_n = n\beta = n\frac{\lambda D}{d}$ where $\lambda$ is the wavelength.

2. For the bright fringes of two wavelengths $\lambda_1 = 480\,\text{nm}$ and $\lambda_2 = 600\,\text{nm}$ to coincide for the first time, their fringe positions must be equal: $n_1 \lambda_1 = n_2 \lambda_2$

3. Substitute the given values: $n_1(480) = n_2(600)$

4. Rearranging, $\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}$

5. For the first coincidence, we need the smallest integers satisfying this ratio: $n_1 = 5, \quad n_2 = 4$

6. This means the $5$-th bright fringe of $480\,\text{nm}$ light coincides with the $4$-th bright fringe of $600\,\text{nm}$ light.

Therefore, the least number of bright fringes of $480\,\text{nm}$ light required is $\boxed{5}$

7. Comparing with the options, this is Option C.