Answer :
Answer:
t = 94.91 nm
Explanation:
given,
wavelength of the light = 522 nm
refractive index of the material = 1.375
we know the equation
c = ν λ
where ν is the frequency of the wave
c is the speed of light
[tex]\nu= \dfrac{c}{\nu\lambda}[/tex]
[tex]\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}[/tex]
ν = 5.75 x 10¹⁴ Hz
the thickness of the coating will be calculated using
[tex]t = \dfrac{\lambda}{4\mu_{material}}[/tex]
[tex]t = \dfrac{522 \times 10^{-9}}{4\times 1.375}[/tex]
t = 94.91 nm
the thickness of the coating will be equal to t = 94.91 nm
The thickness of the coating which is applied to reduce the reflectivity of a pane of glass to light us 94.91 meter.
How to calculate frequency of the wave?
The frequency of the wave can be calculated using the below formula.
[tex]v=\dfrac{c}{\lambda}[/tex]
Here, [tex]c[/tex] is the speed of light and [tex]\lambda[/tex] is the wavelength of the wave.
Given information-
The wavelength of the light is 522 nm.
Index of refraction of material is 1.375.
As the speed of light is [tex]3\times10^8\rm m/s[/tex]. Thus Put the values in the above formula to find the frequency of the light.
[tex]v=\dfrac{3\times10^8 \rm m/s}{522\times10^{-9}\rm m}\\v=5.75\times10^{14}\rm Hz[/tex]
Hence the frequency of the light is [tex]5.75\times10^{14}\rm Hz[/tex].
The thickness of the coating for the above frequency light wave can be find out using the below formula as,
[tex]t=\dfrac{\lambda}{4\times\mu}[/tex]
Here, [tex]\mu[/tex] is the index of refraction of material. Thus put the values,
[tex]t=\dfrac{522\times10^{-9}}{4\times1.375}\\t=94.91\rm m[/tex]
Thus the thickness of the coating which is applied to reduce the reflectivity of a pane of glass to light us 94.91 meter.
Learn more about the frequency of the wave here;
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