Answer:
y = 60λL/πd
Explanation:
To find the distance you can use the expression for the intensity of the fringes in a Young's experiment:
[tex]I=I_ocos^2(\frac{\pi d}{\lambda L}y)[/tex]
d: distance between the slits
L: distance to the screen
y: height of the ring
λ: wavelength of light
For the first maximum you have that I=4I_1, thus, for y=0 (central maximum):
[tex]I=4I_1[/tex]=I_o
to know what is the distance from the central maximum (y=0) you can do y the subject of the formula and replace I=I_1:
[tex]I_1=4I_1cos^2({\frac{\pi d}{\lambda L}}y)\\\\\sqrt{\frac{1}{4}}=cos({\frac{\pi d}{\lambda L}}y)\\\\y=\frac{\lambda L}{\pi d}cos^{-1}(\sqrt{\frac{1}{4}})\\\\y=60\frac{\lambda L}{\pi d}[/tex]
hence, the distance in which he intensity is I1 is 60λL/πd
