How Is the Value of 'm' Determined in Interference Equations?

  • Thread starter Thread starter hyde2042
  • Start date Start date
  • Tags Tags
    Waves
AI Thread Summary
The value of 'm' in interference equations is determined as any integer, including zero, which is often used to simplify calculations. Using 'm=0' is preferred for thinner coatings because it results in lower reflectance at specific wavelengths, maintaining low reflectance across a broader range. The reflectance spectrum indicates that thinner coatings, such as 100 nm, can achieve zero reflectance at a given wavelength, like 600 nm, more effectively than thicker coatings. This approach helps in optimizing the design of optical coatings. Understanding these principles is crucial for accurately determining the thickness of materials in interference applications.
hyde2042
Messages
26
Reaction score
0

Homework Statement


http://i.imgur.com/xrba7.png


Homework Equations


I'm using this equation to find the thickness of the polymer.

[(m+(1/2)) λ = 2nt] (where "n" is the index of refraction and "t" is the thickness )


The Attempt at a Solution



I just want to know how m is determined. I've learned everything else, but I missed the first day on which m value to use and it's always just slipped my mind whenever I'm in class or lab.

Thank you for your time.
 
Physics news on Phys.org
m is any integer, including zero. ehild
 
Thanks, but I'm still unsure as to why "0" is used in the solution instead of "1" or "2".
 
hyde2042 said:
Thanks, but I'm still unsure as to why "0" is used in the solution instead of "1" or "2".

First: it is easier to make a thinner coating then a thick one.

Second: The reflectance depends on the wavelength, this function is about the form R=A+Bcos(4piND/lambda) (N is the refractive index, D the thickness). If the reflectance is zero at a specific wavelength, it stays low in the broadest wavelength range when D has the smallest possible value.
The figure shows the reflectance spectrum of two coatings, one 100 nm, the other 300 nm thick. Both have zero reflectance at 600 nm.


ehild
 

Attachments

  • antirefl.jpg
    antirefl.jpg
    25.2 KB · Views: 399
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top