# Reflection and Film Problem

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1. Nov 24, 2015

### Callix

1. The problem statement, all variables and given/known data
You are designing a thin transparent reflective coating for the front surface of a sheet of glass. The index of refraction of the glass is 1.52 and when it is in use, the coated glass has air on both sides. Because the coating is expensive, you want to use a layer that has the minimum thickness possible, which you determine to be 104nm. What should the index of refraction of the coating be if it must cancel 550nm of light that hits the coated surface at normal incidence?

2. Relevant equations

3. The attempt at a solution
I know that for constructive interference, the equation is $$2d=(m+\frac{1}{2})\frac{\lambda}{n}$$

And for destructive,
$$2d=m\frac{\lambda}{n}$$
However I'm not sure what to do with these.. or if these are exactly the equations I need to use. But these are the notes that I have for thin film.. Any help would be greatly appreciated!

Last edited: Nov 24, 2015
2. Nov 24, 2015

### ehild

The coating must cancel the 550 nm wavelength light, so it must be constructive or destructive interference?
The coating layer must be as thin as possible, so what should be m?

3. Nov 24, 2015

### Callix

Well, it would be destructive since it needs to cancel light.
m should be at a minimum as well, which would be 1, right?

4. Nov 24, 2015

### ehild

Yes, and you know the thickness, so what is the refractive index n?

5. Nov 24, 2015

### Callix

$$n=\frac{550}{2(104)}=2.64$$

6. Nov 25, 2015

### Callix

Hmm.. I was doing some other research and I found some other equations for thin-film interference..

$$\huge \phi=\frac{4\pi n_b t \cos(\theta_i)}{\lambda}+\phi_{r2}-\phi_{r1}$$

Could I make the assumption that since the speed will be greater in air than in the glass that it will experience a pi shift? Therefore, I could simplify the equation to
$$1=\frac{\lambda t}{4n_b}$$

I just found a very similar problem in my book except n for the coating was given and they were solving for the necessary thickness to cancel out 500nm of light. Could I proceed similarly?

Last edited: Nov 25, 2015
7. Nov 25, 2015

### Callix

And since the incidence is normal, cos(0)=1. Well, approximately 0, so cos(~0) = ~1

8. Nov 25, 2015

### ehild

The refractive index of the coating can be either greater or smaller than the refractive index of glass. At the reflection of the air-coating interface, there is a pi shift. At the interface between the layer and the glass, the phase shift is pi if the refractive index of the coating is less then that of the glass, and no phase shift if it is greater than that of the glass. The phase shift during the path of the wave which reflects from the layer-glass interface is (4π/λ)nd.
You assumed the second case, when the net phase shift between the wave reflected from the air-layer interface and that reflected from the layer-glass interface is (4π/λ)nd-π=π, that is, 2d=λ/n.
In case the layer has refractive index is lower that that of glass, the phase shift between the reflected waves is (4π/λ)nd=π, that is, 4d=λ/n.
If you want the thickness of the antireflecting layer you must know what the refractive index of the layer is. It is usually lower than the refractive index of glass.

9. Nov 25, 2015

### Callix

Ah okay, so is my approach correct? And the textbook also notes that if the speed of the first medium is greater than the speed of the second medium, there will also be a phase shift of pi.

10. Nov 25, 2015

### ehild

You had two possibilities for the refractive index. I think you should give both ones.
In general, use the formula in Post #6 and take the phase shifts at the boundaries into account.

11. Nov 25, 2015

### Callix

Ah okay, thank you!