Question

A non-reflective coating that has a thickness of 153 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What wavelength of visible light is most strongly transmitted if it is illuminated perpendicular to its surface?

Answer 1

For most strongly transmitted system, the condition is given by,

2*u*t* cos r = n*lambda --------------------->[1]

'u' is the refractive index of thin film or non-reflective coating = 1.45

't' is the thickness of the non-reflective coating = 153 nm

'r' is the angle of refraction = 0^o (for normal incidence)

'n' is the order of the fringe

'lambda' wave length of the light

plugging in the values in eqn[1], we get

2*u*t* cos r = n*lambda

2*1.45*153*10^-9*1 = n*lambda

n*lambda = 4.437*10^-7 m

for n = 1, lambda = 443.7*10^-9 m = 443.7 nm

for n = 2, lambda = 221.9 nm

for n =3, lambda = 147.9 nm

But the range of visible light is 390-700 nm

Hence, wavelength of the visible light most strongly transmitted is 443.7 nm if it is illuminated perpendicular to its surface.