To solve this exercise it is necessary to use the concepts related to Difference in Phase.
The Difference in phase is given by
[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]
Where
[tex]\delta =[/tex] Horizontal distance between two points
[tex]\lambda =[/tex] Wavelength
From our values we have,
[tex]\lambda = 500nm = 5*10^{-6}m[/tex]
[tex]\theta = 1\°[/tex]
The horizontal distance between this two points would be given for
[tex]\delta = dsin\theta[/tex]
Therefore using the equation we have
[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]
[tex]\Phi = \frac{2\pi(dsin\theta)}{\lambda}[/tex]
[tex]\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}[/tex]
[tex]\Phi= 1.096 rad \approx = 1.1 rad[/tex]
Therefore the correct answer is C.
