A coating of MgF2 (n = 1.38) is deposited on a glass plate (n = 1.50). Light is incident (from air)
onto the plate at angle 60¡ã degrees from the normal to the surface. (Note that this is not nearnormal
incidence and you will have to adjust the equations to allow for the angle)
What is the thinnest MgF2 coating that will minimise reflection of the light at a wavelength
(in air) of 500 nm?
To find the thinnest MgF2 coating that will minimize reflection of light at a wavelength of 500 nm, we can use the concept of interference.
When light passes through a medium with a different refractive index, part of the light is reflected and part of it is transmitted. The reflection and transmission depend on the difference in refractive indices between the two media, as well as the angle of incidence.
In this case, we have light incident from air onto the glass plate coated with MgF2. The first step is to calculate the angle of incidence in the MgF2 layer. We can use Snell's law for this calculation.
n1*sin(theta1) = n2*sin(theta2)
Where:
n1 = refractive index of air = 1 (approx.)
theta1 = angle of incidence in air = 60 degrees
n2 = refractive index of MgF2 = 1.38 (given)
theta2 = angle of incidence in MgF2 (to be calculated)
Using Snell's law, we can rearrange the equation to solve for theta2:
sin(theta2) = (n1/n2)*sin(theta1)
sin(theta2) = (1/1.38)*sin(60 degrees)
sin(theta2) = 0.7241
Taking the inverse sine of this value, we can calculate theta2:
theta2 = arcsin(0.7241)
theta2 = 45.01 degrees (approx.)
Now that we have the angle of incidence in the MgF2 layer, we can calculate the minimum thickness of the MgF2 coating that will minimize reflection.
The condition for minimum reflection occurs when there is a phase difference of λ/2 between the reflected waves from the front and back surfaces of the coating. This results in destructive interference, reducing the overall reflection.
The phase difference is given by:
2*(2π/λ)*(d/λ2) = π
Where:
d = thickness of the MgF2 coating (to be calculated)
λ = wavelength of light in air = 500 nm = 500 * 10^(-9) m
λ2 = wavelength of light in MgF2 = λ/n2 (since the wavelength changes with the refractive index)
Substituting the values into the equation, we can solve for d:
2*(2π/(500 * 10^(-9)))*(d/(500*10^(-9)/1.38)) = π
2*(2π/(500 * 10^(-9)))*(d/(362.32*10^(-9))) = π
(4π/(500 * 10^(-9))) = π/(d/(362.32*10^(-9)))
(500 * 10^(-9))/(4π) = d/(362.32*10^(-9))
d = (500 * 10^(-9))/(4π) * (362.32*10^(-9))
d = 0.363 * 10^(-6) m
The thinnest MgF2 coating that will minimize reflection of the light at a wavelength of 500 nm is approximately 0.363 μm.
To determine the thinnest MgF2 coating that will minimize reflection, we need to calculate the thickness of the coating at which the reflected wave and transmitted wave cancel each other out.
The condition for destructive interference in a thin film is given by the formula:
2nt = (2m+1)λ/2
Where:
- n is the refractive index of the coating material (MgF2) = 1.38
- t is the thickness of the coating
- m is an integer representing the order of the destructive interference
- λ is the wavelength in the medium adjacent to the coating (in this case, air) = 500 nm = 500 x 10^-9 m
Since we want to minimize reflection, we'll consider the order of the first minimum or m = 0.
Now let's solve the equation for t:
2nt = (2m+1)λ/2
2(1.38)(t) = (2(0)+1)(500 x 10^-9) / 2
2.76t = 250 x 10^-9
t = (250 x 10^-9) / 2.76
t = 90.58 x 10^-9 meters
Therefore, the thinnest MgF2 coating that will minimize reflection at a wavelength of 500 nm in air is approximately 90.58 nm.