The Young's double slit interference experiment is performed using light  consisting of  $$480\,nm$$ and  $$600\,nm$$ wavelengths. The least number of the bright fringes of  $$480\,nm$$  light that are  required for the first coincidence with the bright fringes formed by  $$600\,nm$$ light is:

In Young's double slit experiment, the position of the $$n^{th}$$ bright fringe is given by $$y_n = \frac{n \lambda D}{d}$$, where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation. To determine the least number of bright fringes of 480 nm light that will coincide with those of 600 nm light, one sets the positions of the $$n_1^{th}$$ bright fringe of $$\lambda_1 = 480$$ nm and the $$n_2^{th}$$ bright fringe of $$\lambda_2 = 600$$ nm equal. Equating their expressions, $$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$$, simplifies to $$n_1 \lambda_1 = n_2 \lambda_2$$, or $$n_1 \times 480 = n_2 \times 600$$.

This equation can be written as $$\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}$$, meaning the smallest integer values satisfying the ratio are $$n_1 = 5$$ and $$n_2 = 4$$. Verification shows $$5 \times 480 = 2400$$ nm and $$4 \times 600 = 2400$$ nm, confirming that the 5th bright fringe of 480 nm light coincides with the 4th bright fringe of 600 nm light.

Therefore, the least number of bright fringes of 480 nm light required for the first coincidence is 5.

The correct answer is Option 1: 5.