Question

A 1036 nm film with an index of refraction n=2.62 is placed on the surface of glass n=1.52. Light (λ=520.0 nm) falls hits the perpendicular to the surface from air. You want to increase the thickness so the reflected light cancels. What is the minimum thickness of the film that you must add?

Answer 1

Step-by-step explanation:

The ray of light is passing from high refractive index medium to low refractive index medium so condition for cancellation of reflected light is as follows .

2μt = (2n+1) λ/2

where μ is refractive index of the medium , t is thickness , λ is wavelength of light and n is a integer .

Putting n = 10

2x 2.62 x t = 21 x 520 / 2 nm

5.24 t = 5460 nm

t = 1042 nm

Thickness required to be added

= 1042 - 1036 = 6 nm .

Answer 2

`55.64\ \text{nm}`

Step-by-step explanation:

`\lambda` = Wavelength falling on film = 520 nm

n = Refractive index of film = 2.62

T = Thickness of film

m = Order

We have the relation

`2T=(m\lambda)/(n)\\\Rightarrow T=(m\lambda)/(2n)\\\Rightarrow T=(m* 520)/(2* 2.62)\\\Rightarrow T=99.24m`

The thickness should be greater than 1036 nm. This means
`m=11`

`T=99.24* 11=1091.64\ \text{nm}`

Thickness of the film to be added would be

`\Delta T=1091.64-1036=55.64\ \text{nm}`

Thickness of the film to be added is
`55.64\ \text{nm}`.