Young's Double Slit: Phase Shift by π/2?

[S1CkFiSh]

What happens to a light wave when it is reflected off the surface of a material with a higher index of refraction?

a. It reflects with its phase shifted by π.
b. It reflects with its phase shifted by π/2.
c. The phase shift is proportional to the angle of reflection.
d. It reflects with its phase unshifted.
π=pie
From my module it says, Everywhere that the composite wave is bright green is an integer number of wavelengths from both slits. In such places the green from one slit constructively interferes with the green from the other slit (positive plus positive). Everywhere that the composite wave is bright magenta is an integer number plus 1/2 wave from both slits. In such places the magenta from one slit constructively interferes with the magenta from the other slit (negative plus negative). Everywhere that the composite wave is gray is an integer number of waves from one slit, but an integer number plus 1/2 wave from the other slit. In this last case, the green from one slit destructively interferes with the magenta from the other slit (positive plus negative). the pattern of light and dark seen on a screen behind two slits is called a fringe pattern, and is a sure sign of interference and waves.

So I am assuming is it b, it reflects with its phase shifted by π/2? Is that correct or not?
Thank you for reading this, and helping if you can!

 

[lippyka]

Yes, b is the correct answer. When a light wave is reflected off the surface of a material with a higher index of refraction, it reflects with its phase shifted by π/2.

 

[kerimek]

Yes, you are correct. Option b, "It reflects with its phase shifted by π/2," is the correct answer. This is because when a light wave is reflected off a surface with a higher index of refraction, it undergoes a phase shift of π/2. This is due to the change in the speed of the wave as it travels from one medium to another. This phase shift is what causes the interference pattern observed in the Young's double slit experiment.