The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. 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 :-

We need to find the least number of bright fringes of 480 nm light required for the first coincidence with the bright fringes of 600 nm light.

Step 1: Condition for Coincidence

A bright fringe of wavelength $\lambda_1$ coincides with a bright fringe of $\lambda_2$ if their fringe numbers satisfy:

$$n_1 \lambda_1 = n_2 \lambda_2$$

where:

• $n_1$ is the fringe number for $\lambda_1$,
• $n_2$ is the fringe number for $\lambda_2$.

To find the least number of fringes for 480 nm light, we find the Least Common Multiple (LCM) of $\lambda_1$ and $\lambda_2$.

Step 2: Find LCM of Wavelengths

The LCM of 480 nm and 600 nm is:

$$\text{LCM}(480, 600) = 2400 \text{ nm}$$

This means that the first common bright fringe occurs at a path difference of 2400 nm.

Step 3: Find Number of Fringes for 480 nm Light

The number of fringes required for 480 nm light to reach 2400 nm:

$$n_1 = \frac{\text{LCM}}{\lambda_1} = \frac{2400}{480} = 5$$

Final Answer:

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