Answer:[tex]y_1=1.9\ mm[/tex]
Explanation:
Given
slit width [tex]d=1.15\ mm[/tex]
Distance of screen [tex]D=3.53\ m[/tex]
wavelength [tex]\lambda =639\ nm[/tex]
Position of any bright fringe is given by
[tex]y_n=\dfrac{n\lambda D}{d}[/tex]
[tex]y_1=\frac{1\times 639\times 10^{-9}\times 3.53}{1.15\times 10^{-3}}[/tex]
[tex]y_1=0.0019\ m[/tex]
[tex]y_1=1.9\ mm[/tex]
Position of dark fringe is given by
[tex]y_D=\dfrac{(2n+1)\lambda D}{2d}[/tex]
for second dark fringe [tex]n=1[/tex]
[tex]y_D=\dfrac{1.5\times 639\times 10^{-9}\times 3.53}{1.15\times 10^{-3}}[/tex]
[tex]y_D=0.00294\ m\approx 2.94\ mm[/tex]