our formula for double slit interference is:

\(\displaystyle{d}{\sin{{\left({t}\right)}}}={m}{\left({w}\right)}\)

where d is the width of the slits

where t is the angle from the slits

where m is the corresponding fringe from the central fringe

where w is the wavelength

using trig, we can find \(\displaystyle{\sin{{\left({t}\right)}}}\) to be:

\(\displaystyle{\sin{{\left({t}\right)}}}={\frac{{{y}}}{{{L}}}}\)

where y is the distance from central fringe to corresponding fringe

where L is the distance from slit to screen

our formula now becomes:

\(\displaystyle{d}{\left(\frac{{y}}{{L}}\right)}={m}{\left({w}\right)}\)

where both sides represent the path difference:

path difference=d(y/L)

path difference = .150e-3(2e-2/122e-2)

path difference = 2.459e-6 m

In terms of the wavelength:

2.459e-6/561e-9 = 4.4w

\(\displaystyle{d}{\sin{{\left({t}\right)}}}={m}{\left({w}\right)}\)

where d is the width of the slits

where t is the angle from the slits

where m is the corresponding fringe from the central fringe

where w is the wavelength

using trig, we can find \(\displaystyle{\sin{{\left({t}\right)}}}\) to be:

\(\displaystyle{\sin{{\left({t}\right)}}}={\frac{{{y}}}{{{L}}}}\)

where y is the distance from central fringe to corresponding fringe

where L is the distance from slit to screen

our formula now becomes:

\(\displaystyle{d}{\left(\frac{{y}}{{L}}\right)}={m}{\left({w}\right)}\)

where both sides represent the path difference:

path difference=d(y/L)

path difference = .150e-3(2e-2/122e-2)

path difference = 2.459e-6 m

In terms of the wavelength:

2.459e-6/561e-9 = 4.4w