Two slits are 0.800 m from a screen, separated by 0.0720 mm. The two slits...

The intensity at any point can be determined using the relation:
${\displaystyle I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)}$
where ${\displaystyle \varphi }$ is the phase difference corresponding to the given point.
The path difference corresponding to the position y can be given by the following relation:
${\displaystyle \mathrm{△}x=\frac{yd}{D}}$
This path difference for the first minimum is equal to half of the wavelength. Hence, calculate the wavelength using the relation.
${\displaystyle \frac{\lambda }{2}=\frac{(3\times {10}^{-3})(0.072\times {10}^{-3})}{\left(0.800\right)}}$
${\displaystyle \lambda =54\times {10}^{-8}m}$
Now, calculate the path difference corresponding to y=2 mm from the central maximum.
${\displaystyle \mathrm{△}x=\frac{yd}{D}}$
${\displaystyle =\frac{(2\times {10}^{-3})(0.072\times {10}^{-3})}{0.800}}$
${\displaystyle =18\times {10}^{-8}m}$
Now, calculate the phase difference corresponding to the path difference.
${\displaystyle \varphi =\frac{2\pi }{\lambda }\times \mathrm{△}x}$
${\displaystyle =\frac{2\pi }{54\times {10}^{-8}}\times (18\times {10}^{-8})}$
${\displaystyle =\frac{2\pi }{3}}$
The intensity at the point can be determined as follows:
${\displaystyle I=I_{max}{\cos }^2\left(\frac{2\frac{\pi }{3}}{2}\right)}$
${\displaystyle =0.015\frac{W}{m^2}}$