A beam of light consisting of two wavelengths 7000 $$\mathring{A}$$ and 5500 $$\mathring{A}$$ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is $$n \times 10^{-5}$$ m. The value of $$n$$ is ______.

In Young's double slit experiment with wavelengths $$\lambda_1 = 7000 \, \mathring{A}$$ and $$\lambda_2 = 5500 \, \mathring{A}$$, slit separation $$d = 2.5$$ mm, and screen distance $$D = 150$$ cm, find the least distance from the central fringe where bright fringes coincide.

Find fringe widths.

$$\beta_1 = \frac{\lambda_1 D}{d} = \frac{7000 \times 10^{-10} \times 1.5}{2.5 \times 10^{-3}} = 4.2 \times 10^{-4} \text{ m}$$

$$\beta_2 = \frac{\lambda_2 D}{d} = \frac{5500 \times 10^{-10} \times 1.5}{2.5 \times 10^{-3}} = 3.3 \times 10^{-4} \text{ m}$$

Condition for coincidence.

Bright fringes coincide when $$n_1 \beta_1 = n_2 \beta_2$$:

$$n_1 \times 4.2 \times 10^{-4} = n_2 \times 3.3 \times 10^{-4}$$

$$\frac{n_1}{n_2} = \frac{3.3}{4.2} = \frac{33}{42} = \frac{11}{14}$$

Find the least distance.

The smallest integers satisfying this ratio are $$n_1 = 11$$ and $$n_2 = 14$$.

$$y = n_1 \beta_1 = 11 \times 4.2 \times 10^{-4} = 46.2 \times 10^{-4} = 4.62 \times 10^{-3} \text{ m}$$

$$y = 462 \times 10^{-5} \text{ m}$$

Therefore $$n = 462$$.

The correct answer is 462.