Wavelengths / film / reflected light

In summary, the conversation discusses the use of a MgF2 film with a thickness of 1.44 x 10-5 cm to coat a camera lens. The question asks for the three longest wavelengths that are intensified in the reflected light, with the given equations for constructive and destructive interference. It is determined that the equation for constructive interference is incorrect and the correct values can be found by calculating lambda for m=1, 2, 3. The conversation also touches on the importance of understanding the derivation of the formula for thin film interference.
  • #1
Color_of_Cyan
386
0

Homework Statement



A film of MgF2 (n = 1.38) having thickness 1.44 x 10-5 cm is used to coat a camera lens.
What are the three longest wavelengths that are intensified in the reflected light? (Enter your answers from smallest to largest.)


Homework Equations


λn = λ/n

n = index of refrection

for constructive interference:
2t = (m + 1/2)λn
2nt = (m + 1/2)λ

destructive interference:
2nt = mλ

The Attempt at a Solution


Honestly have no idea what the question is asking...

not sure how to combine them, not sure how to find 3 biggest wavelengths

not sure if it is constructive interference or not soo

2nt = (m + 1/2)λ

2(1.38m)(0.000000144m) = (0 + 1/2)λ

(2)(1.38m)(0.000000144m) / (1/2) = λ

λ = 795 nano meters (for m = 0)

^^ Yet says it is wrong

All I can think of is to change m to 1, 2, 3 etc but don't think it works.



Suggestions??
 
Physics news on Phys.org
  • #2
Color_of_Cyan said:
for constructive interference:
2t = (m + 1/2)λn
2nt = (m + 1/2)λ
This assumes that the index of refraction of the lens is less than that of the coating. What's the index of refraction of the lens?
 
  • #3
It actually doesn't say it just says it for the coating.

I don't know how to find it either, any hints??
 
Last edited:
  • #4
I would assume that the index of refraction of the lens (presumably glass) is greater than that of the coating. How would that change your approach?
 
  • #5
I don't really understand what you are trying to imply...

Would I have to use the index of refraction of the lens instead?

( I read that for crown glass, it would be n = 1.52 instead. Is this supposed to be "given" to you ? )
 
  • #6
Realize that there is a λ/2 phase shift upon reflection from a surface with higher index of refraction. So it matters. As I said before, just assume that the index of refraction of the lens is greater than that of the coating. That's all you need to know.
 
  • #7
Doc Al said:
Realize that there is a λ/2 phase shift upon reflection from a surface with higher index of refraction. So it matters. As I said before, just assume that the index of refraction of the lens is greater than that of the coating. That's all you need to know.
I can't see how that is supposed to help :uhh:

I seriously can't find any way how to incorporate the phase shift in the problem.


Also is it right that you DO get each of the big wavelengths with m equal to 0, 1, 2, etc?
 
  • #8
Color_of_Cyan said:
for constructive interference:
2t = (m + 1/2)λn
2nt = (m + 1/2)λ

destructive interference:
2nt = mλ

You cited wrong conditions for constructive and destructive interference in the reflected light.
2nt = (m + 1/2)λ refers to destructive interference (minimum reflectance) and 2nt = mλ for the constructive one (maximum reflectance). The problem asks the wavelengths where constructive interference occurs.
Calculate lambda for m=1, 2, 3.

ehild
 
  • #9
Color_of_Cyan said:
I can't see how that is supposed to help :uhh:

I seriously can't find any way how to incorporate the phase shift in the problem.
You need to review how the formula for thin film interference is derived. Whether there is a phase shift on reflection at each surface depends on the relative index of refraction. Assuming that the index of refraction of the coating is less than that of the lens (a reasonable assumption), then your equation for constructive interference is wrong. (As ehild points out, your version describes the condition for destructive interference. But do not simply copy the formula--understand how it is derived and then you'll be good to go no matter what kind of thin film is used.)
Also is it right that you DO get each of the big wavelengths with m equal to 0, 1, 2, etc?
Yes, that's the right idea. But first get the correct formula.
 
  • #10
ehild said:
You cited wrong conditions for constructive and destructive interference in the reflected light.
2nt = (m + 1/2)λ refers to destructive interference (minimum reflectance) and 2nt = mλ for the constructive one (maximum reflectance). The problem asks the wavelengths where constructive interference occurs.
Calculate lambda for m=1, 2, 3.

ehild
Thank you, I got it now, and I also got another problem right that I had wrong because of you saying this too !

Doc Al said:
You need to review how the formula for thin film interference is derived. Whether there is a phase shift on reflection at each surface depends on the relative index of refraction. Assuming that the index of refraction of the coating is less than that of the lens (a reasonable assumption), then your equation for constructive interference is wrong. (As ehild points out, your version describes the condition for destructive interference. But do not simply copy the formula--understand how it is derived and then you'll be good to go no matter what kind of thin film is used.)
Blah, I blame my book for not seeing that. Then again I sort of glanced over it too.
 
Last edited:

1. What is a wavelength?

A wavelength is the distance between two consecutive peaks or troughs of a wave. In the context of light, it refers to the distance between two consecutive crests or troughs of an electromagnetic wave.

2. How does the wavelength of light affect the color of objects?

The wavelength of light determines the color of an object by determining which wavelengths of light are reflected and which are absorbed. Objects appear a certain color because they reflect that particular wavelength of light and absorb all others.

3. What is film sensitivity to different wavelengths of light?

Film sensitivity refers to how well a particular type of film can capture different wavelengths of light. Some films are more sensitive to certain wavelengths, which can result in variations in color and contrast in the final image.

4. How does reflected light affect the appearance of objects?

Reflected light plays a crucial role in how we perceive the color and texture of objects. The wavelengths of light that are reflected off an object determine its color, and the direction and intensity of the reflected light contribute to its perceived texture and shine.

5. How do different wavelengths of light affect how we see the world?

Our eyes are only sensitive to a small portion of the electromagnetic spectrum, known as visible light. Different wavelengths of light within this spectrum affect how we see the world by influencing the colors we perceive, as well as the brightness and contrast of objects.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
3
Views
843
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
8K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
903
  • Introductory Physics Homework Help
Replies
1
Views
759
Back
Top