Two flat slides of glass are separated at one edge by a thin wire, as shown below. The top surface of the upper slide and the bottom surface of the lower slide have special coatings on them so that they reflect no light. The system is illuminated with light of wavelength 550 nm. Looking down from above you see 40 interference minima. What is the diameter of the wire? Assume that the last minima occurs at the right edge where the wire is placed.
Do your own CAPA.
Well this is quite trivial no?
sooooooooooooooooooooooo easy
use halbadier's rule....... simple
Got it, thanks!
what on earth is: halbadier's rule?
what the **** is halbadier's rule? lol
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what the *beep* is going on with thie question? I'm seriously lost at this point ....
To find the diameter of the wire, we can use the concept of interference of light waves. The interference pattern is formed due to the waves reflecting off the top and bottom surfaces of the glass slides and interfering with each other.
Given that there are 40 interference minima, we can determine the path difference between the two interfering waves using the following formula:
Path Difference = (m + 0.5) * λ
where m is the order of the interference minima (in this case, the number of minima is given as 40), and λ is the wavelength of light (550 nm = 550 × 10^(-9) m).
Since the wire separates the two slides, it causes an additional path difference of 2t, where t is the thickness of the wire.
Now, let's set up the equations to find the diameter of the wire:
Path Difference = 2t
(m + 0.5) * λ = 2t
Since we want to find the diameter of the wire, we need to express t in terms of the diameter (d):
t = d/2
Substituting this into the equation:
(m + 0.5) * λ = 2 * (d/2)
(m + 0.5) * λ = d
Now, we can solve for the diameter by rearranging the equation:
d = (m + 0.5) * λ
Substituting the given values:
d = (40 + 0.5) * 550 × 10^(-9) m
d = 22.75 * 550 × 10^(-9) m
d ≈ 12.51 × 10^(-6) m
Therefore, the diameter of the wire is approximately 12.51 micrometers.