Answer:
y(m=+1,-1)=3.36mm
Explanation:
We have to take into account the expression for the position of the fringes
![y=\frac{m\lambda D}{d}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bm%5Clambda%20D%7D%7Bd%7D)
Where lambda is the wavelength of the light, D is the distance to the screen, m is the order of the fringe and d is the distance between slits.
By replacing we have
![y=\frac{(1)(600*10^{-9}m)(1.4m)}{0.25*10^{-3}m}=3.36*10^{-3}m](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%281%29%28600%2A10%5E%7B-9%7Dm%29%281.4m%29%7D%7B0.25%2A10%5E%7B-3%7Dm%7D%3D3.36%2A10%5E%7B-3%7Dm)
There is a distance of 3.36mm to the secon maximum in the screen.
HOPE THIS HELPS!!