Question

a) What is the minimum index of refraction of a clear material if a minimum thickness of 122 nm , when laid on glass, is needed to reduce reflection to nearly zero when light of 675 nm is incident normally upon it? Assume that the film has an index less than that of the glass.

I got n_min=2.76 and it is saying incorrect.. used the equation 2t=m λ_n not sure what I am doing wrong

b) What is the main index of refraction of the material if the film has an idea greater than that of the glass?

Answer 1

Assume that, the film has an index less than that of the glass.

(a) The minimum index of refraction of a clear material which will be given as :

using an equation,    t = lamda / 4 n_min                                                   { eq.1 }

where, t = thickness of material = 122 nm

lamda = wavelength of light = 675 nm

n_min = minimum refractive index of material

inserting the values in above eq.

(122 nm) = (675 nm) / 4 n_min

n_min = (675 nm) / 4 (122 nm)

n_min = 1.38

(b) The main index of refraction of material, if the film has an idea greater than that of the glass which is given as :

using an equation,    n_c = sqrt (n_g)                                                             { eq.2 }

where, n_g = refractive index of glass = 1.5

inserting the value of n_g in eq.2,

n_c = sqrt (1.5)

n_c = 1.22