In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is _____ nm.

In Young's double slit experiment, the fringe width (separation between consecutive bright fringes) is given by:

$$\beta = \frac{\lambda D}{d}$$

where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation.

Given data:

For the first light: $$\lambda_1 = 560$$ nm, $$\beta_1 = 7.2$$ mm

For the second light: $$\beta_2 = 8.1$$ mm, $$\lambda_2 = ?$$

Since the experimental setup (D and d) remains the same:

$$\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}$$

Solving for $$\lambda_2$$:

$$\lambda_2 = \lambda_1 \times \frac{\beta_2}{\beta_1}$$

$$\lambda_2 = 560 \times \frac{8.1}{7.2}$$

$$\lambda_2 = 560 \times 1.125$$

$$\lambda_2 = 630 \text{ nm}$$

Therefore, the wavelength of the second light is 630 nm.