Assume the wavelength of visible light lie between 400nm and 710nm. 1. how many
ID: 2124767 • Letter: A
Question
Assume the wavelength of visible light lie between 400nm and 710nm.
1. how many phase changes? I hae put 1.
2. which visible wavelengths interfere constructively?
do you use lambda_const=2nt/m+1/2 and use m=1,2,3,4 to find the wavelengths where visible bc i got at m=3,4 with wavelngths 623 and 484.4 nm
3. visible wavelenths interfere destructively?
4. maximu thickness the layer can have so that no visible wavelenths interfere constuctively?
Please explain with great detail will give max points for a thorough explanation
Explanation / Answer
Given refractive index of the coating material = 2.18
Given the thickness of the coating = 500 nm
Wavelength of visible ligh = 400 nm to 710 nm
a) Phase change occurs when wave get reflected while it goes from lower refractive index to higher refractive index. It goes total of 1 times - 1st when it enters air to coating layer. Hence phase change occurs only once.
b) For constructive interference
2*n(Coating)*d*cos(?) = n?
Assuming normal incidence
2*2.18*500*10^-9 = n?
? = (2*2.18*500*10^-9)/n
? = (2180/n) nm
Now n = 1, 2, 3 ....
Least
400 = 2180/n
n = 5.45
For maximum wavelength
710 = 2180/n
n = 3.07
So we have only 2 integers between 3.07 and 5.45, those are 4 and 5
so we have wavelengths of light which constructively interfere are
2180/4 nm and 2180/5 nm that is 545 nm and 436 nm
Constructively interfere are 436 nm and 545 nm.
c) For destructive interference
2*n(coating)*d*cos(?) = (n-(1/2))?
2*2.18*500 = (n-(1/2))?
Similarly solving like in previous case
? = 2180/(n-(1/2))
400 = 2180/(n-(1/2))
n-(1/2) = 5.45
n = 5.95
710 = 2180/(n-(1/2))
n = 3.57
Again we have 2 integers 4 and 5 in between
the wavelengths corresponding are
2180/(3.5) and 2180/(4.5) nm
622.85 nm and 484.44 nm