Light - Thin Film interference

In summary, a thin film of MgF2 with an index of refraction of 1.38 coats a piece of glass, and constructive interference is observed for the reflection of light with wavelengths of 500 nm and 625 nm. Using the equation t = lambda/2n *(m + .5), the thinnest film for which this can occur is 906 nm.
  • #1
Foxhound101
52
0

Homework Statement


A thin film of MgF2 with n = 1.38 coats a piece of glass. Constructive interference is observed for the reflection of light with wavelengths of 500 nm and 625 nm.

Part A -
What is the thinnest film for which this can occur?

Homework Equations


t = lambda/2n *(m + .5)
and/or
lambda=(2*n*t)/(m+.5)


The Attempt at a Solution


Not sure what I am supposed to do, because of the two wavelengths. I know what to do for a single wavelength...

So this is my work for solving the problem with each wavelength

m=0,1,2,3,...

500 = 2(1.38)(t)/(m+.5)

500(0+.5) = 2.76(t)

t = 90.6

and

625(0+.5) = 2.76(t)
t = 113

Now that I have done this, I am not sure how to proceed. Any help is appreciated.
 
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  • #2
Hi Foxhound101,

Foxhound101 said:

Homework Statement


A thin film of MgF2 with n = 1.38 coats a piece of glass. Constructive interference is observed for the reflection of light with wavelengths of 500 nm and 625 nm.

Part A -
What is the thinnest film for which this can occur?

Homework Equations


t = lambda/2n *(m + .5)
and/or
lambda=(2*n*t)/(m+.5)

I don't think these are correct. I believe these equations would be for destructive interference; isn't the index of refraction of glass greater than 1.38?

The Attempt at a Solution


Not sure what I am supposed to do, because of the two wavelengths. I know what to do for a single wavelength...

So this is my work for solving the problem with each wavelength

m=0,1,2,3,...

500 = 2(1.38)(t)/(m+.5)

500(0+.5) = 2.76(t)

t = 90.6

and

625(0+.5) = 2.76(t)
t = 113

Now that I have done this, I am not sure how to proceed. Any help is appreciated.

By setting m=0, you have found a minimum thickness for each wavelength (for destructive interference with the equations you are using). But in this problem you don't want the minimum for each; you want a thickness t that is true for both wavlengths. So use m1 for one wavelength and m2 for another wavelength.

After eliminating t, you can find a relationship between the two m values. What do you get?
 
  • #3
Thanks for the help, alphysicist

<i>isn't the index of refraction of glass greater than 1.38?</i> Index of refraction in glass is greater than 1.38...the value is sometimes 1.55

So...another formula I have is t=(m*lambda)/(2n)

So...
lambda = 500

t=(1*500)/(2*1.38)
t= 181

t= (2*500)/(2*1.38)
t= 362

lambda = 625
t = (1*625)/(2*1.38)
t = 226

t = (2*625)/(2*1.38)
t = 453
 
  • #4
So the relationship between these values...

181/226 = .8

362/453 = .8

Interesting...so what do I do, now that I know this relationship?
 
  • #5
Foxhound101 said:
Thanks for the help, alphysicist

<i>isn't the index of refraction of glass greater than 1.38?</i> Index of refraction in glass is greater than 1.38...the value is sometimes 1.55

So...another formula I have is t=(m*lambda)/(2n)

That's the right formula for this problem. (If, for example, the thin film was on something that with an index of refraction less than 1.38, then your original problem would have been correct for constructive interference.)

So...
lambda = 500

t=(1*500)/(2*1.38)
t= 181

t= (2*500)/(2*1.38)
t= 362

lambda = 625
t = (1*625)/(2*1.38)
t = 226

t = (2*625)/(2*1.38)
t = 453


I don't think this is the way you want to do it. You are setting the m's to be specific numbers; but in this problem they are unknown. Just leave them as m1 and m2:

t=(m1*500)/(2*1.38)

t = (m2*625)/(2*1.38)

So you have these two equations; now eliminate t to find out how the two m values are related. For example, if you end up with:

m1 = 1.7 m2

then you know that m1=17, m2=10 would be the smallest integer values that m1 and m2 could have that still obey that relationship.



(The ratios you found in your last post are related to this, but I think it rather hides the process.)
 
  • #6
t=(m*500)/2.76
t=(m*625)/2.76

(m1*500)/2.76 = (m2*625)/2.76
m1 * 500 = m2 * 625

m1 = .8m2

m1 = 10 and m2 = 8
 
  • #7
Plugging those m's in...

t = (10*500)/(2.76)
t = 1812

t = (8*625)/(2.76)
t = 1812

Does this seem like a good answer? I typed 1810 into the program and it told me I was still wrong.
 
  • #8
Foxhound101 said:
t=(m*500)/2.76
t=(m*625)/2.76

(m1*500)/2.76 = (m2*625)/2.76
m1 * 500 = m2 * 625

m1 = .8m2

m1 = 10 and m2 = 8

I think you have those backwards (m1 is 8 and m2 is 10).

However, those are not the smallest integers that satisfy the relationship. (Rewrite 0.8 as a fraction and then simplify.)

Once you get the set of smallest integers, what do you get for the minimum thickness?
 
  • #9
If I change m1 and m2, then they won't equal each other.

If I leave m1 = 5 and m2 = 4 then I get...

(5*500)/2.76
t = 906

t=(4*625)/2.76
t = 906

This seems like a decent answer...answers for other problems were t=1000ish, so this is decent.

Does this seem like a good answer now?
 
  • #10
Foxhound101 said:
If I change m1 and m2, then they won't equal each other.

Okay, I see what happened. You wrote

m1 = .8m2

a few posts ago but you meant

m2 = .8m1

So your numbers for m1 and m2 are fine.

If I leave m1 = 5 and m2 = 4 then I get...

(5*500)/2.76
t = 906

t=(4*625)/2.76
t = 906

This seems like a decent answer...answers for other problems were t=1000ish, so this is decent.

Does this seem like a good answer now?

That looks right to me (if they want the answer in nanometers).
 
  • #11
That was the correct answer. Thank you for your help.
 
  • #12
Sure, glad to help!
 

FAQ: Light - Thin Film interference

What is thin film interference?

Thin film interference is a phenomenon that occurs when light waves reflect off of two surfaces that are very close together, creating a pattern of constructive and destructive interference. This results in certain wavelengths of light being amplified and others being cancelled out, producing a spectrum of colors.

What are some examples of thin film interference in everyday life?

Thin film interference can be observed in many everyday objects, such as soap bubbles, oil slicks, and the colors on the surface of a CD or DVD. It is also used in anti-reflective coatings on glasses and camera lenses, as well as in the production of holograms.

How is thin film interference related to the thickness of the film?

The thickness of the film determines the wavelength of light that will be amplified or cancelled out. When the thickness of the film is an integer multiple of the wavelength of light, there will be constructive interference and a bright color will be observed. When the thickness is a half-integer multiple, there will be destructive interference and a dark color will be observed.

What factors affect the appearance of thin film interference?

The appearance of thin film interference can be affected by the angle of incidence of light, the refractive index of the film material, and the wavelength of light. These factors can alter the path and phase of the light waves, resulting in different patterns of interference.

How is thin film interference used in scientific research?

Thin film interference is used in various fields of research, such as optics, materials science, and chemistry. It can be used to measure the thickness of films, study the properties of materials, and analyze the composition of substances. It is also used in the development of new technologies, such as solar cells and optical coatings.

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