Can someone please help me with this? Thanks in advance/ What is the minimum thi
ID: 1780452 • Letter: C
Question
Can someone please help me with this? Thanks in advance/
What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 431 nm light? The index of refraction e coating material is 1.31 and the index of the glass is 1.56 1.412 E2 nm You must take into account the 180 degree phase-jump on reflection of light from a medium with a lower index of refraction to a medium of er index of refraction. Also remember that the wavelength of light changes in a medium. The antireflective coating should be half of the nimum length you determine is necessary for destructive interference to occuExplanation / Answer
path difference = 2t
for minimumreflection path difference = lambda/2 = lambda_0/(2n)
lamba_0 = wavelength in air
n = refractive index of coating
2t = lamnda_0/(2n)
2*t = 431*10^-9/(2*1.31)
thickness = 82.2 nm <<<<<============ANSWER