In Young's double slits experiment fringe width is $\beta$. Intensity at a point on the screen does not change when one of the slit is closed or opened, then the minimum distance of that point from the central maximum is 

Intensity at any point on the screen is $I= I_0{\cos }^2\left(\frac{\delta }{2}\right)$ where $\delta$ is the phase difference between the waves meeting at that point. When one of the slit is closed intensity at any point is due to one source i.e. $\frac{I_0}{4}$

given $\frac{I_0}{4}=I_0{\cos }^2\left(\frac{\delta }{2}\right)$ solving we get $\delta =\frac{2\pi }{3}$ i.e. corresponding path difference is $\frac{\lambda }{3}$

for $\lambda$ path difference distance is $\beta$ for $\frac{\lambda }{3}$it is $\frac{\beta }{3}$