Film Thickness for Minimum Reflection of Monochromatic Light: How to Calculate?

In summary, the minimum film thickness that will result in minimum reflection of the monochromatic light is λ/4n1. The correct equation to use is 2t = (m+0.5)λ/n, where m is an integer and n is the refraction index of the material the light is traveling through. The question involves destructive interference, meaning there is a phase shift of pi between the air and the material, as well as between the two materials. Therefore, m=0 is not a possible solution. The denominator should be n1, since the light does not enter the material with index of refraction n2.
  • #1
Erickly
14
2

Homework Statement


Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n2 is larger than n1, what minimum film thickness will result in minimum reflection of this light?

Multiple choice:
A λ/n1
B λ/(4n1)
C λ/n2
D λ/2
E λ
F λ/(4n2)
G λ/(2n1)
H λ/4
I λ/(2n2)
J The correct response is not shown above.

Homework Equations


2t = m*λ/n

2t = (m+0.5)*λ/n

where m is an integer, n is the refraction index of light, and t is thickness.

The Attempt at a Solution


So I believe the wave is destructive because there are 2 phase changes. One when light enters it and the second when it hits passes the coating. Therefore the equation is the second one. I reasoned M must be zero because otherwise I would get a number in the numerator when there is not one in the multiple choice. So if we dived the 2 we get t = λ/4n(something) I don't know if the n is supposed to be 1 or 2.

Any help is greatly appreciated!
 
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  • #2
Erickly said:

Homework Statement


Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n2 is larger than n1, what minimum film thickness will result in minimum reflection of this light?

Multiple choice:
A λ/n1
B λ/(4n1)
C λ/n2
D λ/2
E λ
F λ/(4n2)
G λ/(2n1)
H λ/4
I λ/(2n2)
J The correct response is not shown above.

Homework Equations


2t = m*λ/n

2t = (m+0.5)*λ/n

where m is an integer, n is the refraction index of light, and t is thickness.

The Attempt at a Solution


So I believe the wave is destructive because there are 2 phase changes. One when light enters it and the second when it hits passes the coating. Therefore the equation is the second one. I reasoned M must be zero because otherwise I would get a number in the numerator when there is not one in the multiple choice. So if we dived the 2 we get t = λ/4n(something) I don't know if the n is supposed to be 1 or 2.
Any help is greatly appreciated!
You are on the right track. The second equation is the right one. What is wrong with m=0?
As for whether the denominator should be n1 or n2: What is the optical path length for a ray that goes through n1, reflects at the n1-n2 boundary, then passes through n1 again? That should answer if it's n1 or n2 ...
 
  • #3
rude man said:
You are on the right track. The second equation is the right one. What is wrong with m=0?
As for whether the denominator should be n1 or n2: What is the optical path length for a ray that goes through n1, reflects at the n1-n2 boundary, then passes through n1 again? That should answer if it's n1 or n2 ...
So I guess since it goes through a phase change m cannot be zero (maybe m = 2?)? Also would the n be n1 because it never enters n2? (assuming it has 1 phase change when entering n1 and then bounces off n2 without entering it)
 
  • #4
Erickly said:
So I guess since it goes through a phase change m cannot be zero (maybe m = 2?)? Also would the n be n1 because it never enters n2? (assuming it has 1 phase change when entering n1 and then bounces off n2 without entering it)

Answer to second question - yes.

Again - why not m=0? You say there's a "phase shift" (meaning a phase shift of pi) between n1 and n2, which is correct. But what about a phase shift between the air and n1? What if both beams get a phase shift of pi each, what would that say about the net phase shift between the two beams due to reflection alone? And what does the total shift between the two beams have to be to get destructive interference?

If you're going to understand what's going on you need to understand why you picked the second equation. As it is you just lucked out. ( I really shouldn't have confirmed that it's the right one in the first place.)
 
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  • #5
rude man said:
Answer to second question - yes.

Again - why not m=0? You say there a "phase shift" (meaning a phase shift of pi) between n1 and n2, which is correct. But what about a phase shift between the air and n1? What if both beams get a phase shift of pi each, what would that say about the net phase shift between the two beams due to reflection alone? And what does the total shift between the two beams have to be to get destructive interference?

If you're going to understand what's going on you need to understand why you picked the second equation. As it is you just lucked out. ( I really shouldn't have confirmed that it's the right one in the first place.)

So I guess my problem is I don't really understand what m is. I know its an integer and thought is it the number of times light passes through a medium that causes part of the ray to reflect. That is why I thought it is 2.

As for destructive, I know what it means ( the waves, when added together cancel parts of each other out). I thought this is destructive because n1 < n2 (similar to another example problem I had in lecture ).

Do I have the right idea here?
 
  • #6
Erickly said:
So I guess my problem is I don't really understand what m is. I know its an integer and thought is it the number of times light passes through a medium that causes part of the ray to reflect. That is why I thought it is 2.
m is any positive integer including zero. It's the number of wavelengths of total optical path difference, corrected if necessary by reflection of half a wavelength. But you should not worry about that. To you m is just any positive integer. Any positive integer value of m makes the equation correct.
As for destructive, I know what it means ( the waves, when added together cancel parts of each other out). I thought this is destructive because n1 < n2 (similar to another example problem I had in lecture ).
They don't cancel parts of each other out, they cancel all of each other out.

Yes, going from n1 to n2 means destructive interference by itself. But there is more to the problem than that. There is the phase shift due to travel inside n1 and there may be a second pi phase shift between the air and n1 which you have to decide is the case or not.
 
  • #7
rude man said:
m is any positive integer including zero. It's the number of wavelengths of total optical path difference, corrected if necessary by reflection of half a wavelength. But you should not worry about that. To you m is just any positive integer. Any positive integer value of m makes the equation correct.
They don't cancel parts of each other out, they cancel all of each other out.

Yes, going from n1 to n2 means destructive interference by itself. But there is more to the problem than that. There is the phase shift due to travel inside n1 and there may be a second pi phase shift between the air and n1 which you have to decide is the case or not.

Ohhh the first part makes more sense now, its like a unit circle and adding 2pi correct? if you take the sin of 60deg and sin of 60+360deg its the same value. Is that the correct way of thinking about m?

As for the second part I believe there is a phase change between air and n1 because n_air =/= n1. If that is true then what do I do with that information?
 
  • #8
Erickly said:
Ohhh the first part makes more sense now, its like a unit circle and adding 2pi correct? if you take the sin of 60deg and sin of 60+360deg its the same value. Is that the correct way of thinking about m?
Most definitely, yes.
As for the second part I believe there is a phase change between air and n1 because n_air =/= n1. If that is true then what do I do with that information?
How about combining that with the other reflection (n1 to n2) and the optical path length difference to come up with destructive interference between the two beams?
 
  • #9
Well would it give me this relationship: 2t = ( m + 0.5 ) λi / n where lambda i is wave length in air and therefore n must be n1? That way I relate air to n1. Since n1 < n2 perhaps it bounces off n2 but doesn't pass it, therefore I do not use it in the equation?

Is that what you meant? I'm sorry its taking me so long to understand this.
 
  • #10
Erickly said:
Well would it give me this relationship: 2t = ( m + 0.5 ) λi / n where lambda i is wave length in air and therefore n must be n1? That way I relate air to n1. Since n1 < n2 perhaps it bounces off n2 but doesn't pass it, therefore I do not use it in the equation?

Is that what you meant? I'm sorry its taking me so long to understand this.
That's not all I meant but yes, the rays don't penetrtate into n2 so the only thing to worry about there is the phase shift at the n1 - n2 interface. n2 does not enter into the calculations other than the pi phase shift.
 
  • #11
rude man said:
That's not all I meant but yes, the rays don't penetrtate into n2 so the only thing to worry about there is the phase shift at the n1 - n2 interface. n2 does not enter into the calculations other than the pi phase shift.

Okay so if we plug in values, ( m = 0 // N = n1 ) t will be Lambda over (4*n1)?
 
  • #12
Erickly said:
Okay so if we plug in values, ( m = 0 // N = n1 ) t will be Lambda over (4*n1)?
I'd put my money on that! :smile:
Congrats for not letting go!
 
  • #13
rude man said:
I'd put my money on that! :smile:
Congrats for not letting go!

Thank you so much for all your help! I'm very grateful, hopefully I can pass it forward someday!
 

Related to Film Thickness for Minimum Reflection of Monochromatic Light: How to Calculate?

1.

What is wave optics interference?

Wave optics interference is a phenomenon in which two or more waves interact with each other, resulting in a new pattern of wave amplitudes and intensities. This can occur when two light waves, sound waves, or any other type of wave overlap in space and time.

2.

How does wave optics interference occur?

Wave optics interference occurs when two or more waves of the same frequency and wavelength meet at a point in space. At this point, the waves will either reinforce each other or cancel each other out, depending on their relative phase difference.

3.

What is the difference between constructive and destructive interference in wave optics?

Constructive interference occurs when two waves with the same frequency and wavelength are in phase, meaning their peaks and troughs align, resulting in a larger amplitude. Destructive interference occurs when two waves are out of phase, resulting in a cancellation of the waves and a smaller amplitude.

4.

How is the intensity of interference patterns calculated?

The intensity of interference patterns can be calculated by squaring the amplitudes of the interfering waves and adding them together. This is known as the principle of superposition and is used to determine the overall intensity of the resulting wave pattern.

5.

What are some practical applications of wave optics interference?

Wave optics interference has many practical applications, including in the fields of optics, acoustics, and telecommunications. It is used in technologies such as holography, interferometry, and fiber optic communication systems. It is also used in medical imaging techniques, such as ultrasound and MRI, to produce detailed images of internal structures. Additionally, wave optics interference is essential in the study of light and its properties, which has led to advancements in fields such as astronomy, meteorology, and physics.

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