In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index $$\frac{7}{5}$$, is $$\frac{1}{\alpha}$$. The value of $$\alpha$$ is ______.

The angular width of the fringe in air is $$\theta = 0.35°$$, the distance to the screen is $$D = 2$$ m, the wavelength is $$\lambda = 450$$ nm, and the refractive index of the medium is $$\mu = \frac{7}{5}$$.

In Young's double slit experiment, the angular width of a fringe is given by $$\theta = \frac{\lambda}{d}$$ where $$d$$ is the slit separation. When the system is immersed in a medium of refractive index $$\mu$$, the effective wavelength becomes $$\lambda' = \frac{\lambda}{\mu}$$, so the new angular width is $$\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}.$$

Substituting the values, $$\theta' = \frac{0.35°}{\frac{7}{5}} = 0.35° \times \frac{5}{7} = 0.25° = \frac{1}{4}°.$$

Since the angular width is $$\frac{1}{\alpha}$$ degrees, we have $$\frac{1}{\alpha} = \frac{1}{4} \quad\Longrightarrow\quad \alpha = 4.$$

Hence, the value of $$\alpha$$ is 4.