In a Young's double slit experiment, an angular width of the fringe...

Angular fringe width is given by $\theta = \dfrac{\lambda}{D}$, where, $D$ is the distance between the source and the screen and $\lambda$ is the wavelength of the light.

We have, $\dfrac{{\theta}_{1}}{{\theta}_{2}} = \dfrac{{\lambda}_{1}}{{\lambda}_{2}}$

Then, ${\theta}_{2} = \dfrac{{0.35}^{∘}}{450nm} \times \dfrac{450nm}{\dfrac{7}{5}} = {0.25}^{∘} = \dfrac{1}{4} \left(∵ {\lambda}_{2} = \dfrac{{\lambda}_{1}}{\mu}\right)</p><p>Thus, value$ of \alpha = 4$.</p>$