5. The surface of a glass plate (index of refraction n3 = 1.50) is coated with a transparent thin film (index of refraction n2 = 1.25). A beam of monochromatic light of wavelength 6.0 X 10^-7 meter traveling in air (index of refraction n1 = 1.00) is incident normally on surface S1 as shown above. The beam is partially transmitted and partially reflected.
(a) Calculate the frequency of the light.
(b) Calculate the wavelength of the light in the thin film.
The beam of light in the film is then partially reflected and partially transmitted at surface S2·
(c) Calculate the minimum thickness d1 of the film such that the resultant intensity of the light reflected back into the air is a minimum.
(d) Calculate the minimum nonzero thickness d2 of the film such that the resultant intensity of the light reflected back into the air Is a maximum.
I need help on this plzzz
(a) To calculate the frequency of the light, we can use the equation:
c = λν
where c is the speed of light, λ is the wavelength, and ν is the frequency.
We are given the wavelength λ of the light as 6.0 X 10^-7 meter. We can substitute this value into the equation and solve for the frequency ν:
c = (6.0 X 10^-7 m)ν
Rearranging the equation to solve for ν:
ν = c / (6.0 X 10^-7 m)
The speed of light is approximately 3.00 X 10^8 m/s, so substituting this value:
ν = (3.00 X 10^8 m/s) / (6.0 X 10^-7 m)
Simplifying the expression:
ν = 5.00 X 10^14 Hz
Therefore, the frequency of the light is 5.00 X 10^14 Hz.
(b) To calculate the wavelength of the light in the thin film, we can use the equation:
λ2 = λ1 / n2
where λ2 is the wavelength in the thin film, λ1 is the wavelength in air, and n2 is the refractive index of the thin film.
We are given the wavelength in air λ1 as 6.0 X 10^-7 meter and the refractive index of the thin film n2 as 1.25. We can substitute these values into the equation and solve for λ2:
λ2 = (6.0 X 10^-7 m) / 1.25
Simplifying the expression:
λ2 = 4.8 X 10^-7 m
Therefore, the wavelength of the light in the thin film is 4.8 X 10^-7 meter.
(c) To calculate the minimum thickness d1 of the film such that the resultant intensity of the light reflected back into the air is a minimum, we need to consider the conditions for destructive interference.
Destructive interference occurs when the path difference between the reflected and transmitted rays is equal to an odd multiple of half the wavelength in the film (λ2/2).
The path difference can be calculated as:
Δx = 2n2d1
where Δx is the path difference, n2 is the refractive index of the thin film, and d1 is the thickness of the film.
Since we want to minimize the reflected intensity, we want the path difference to be equal to λ2/2.
Setting up the equation:
λ2/2 = 2n2d1
Solving for d1:
d1 = λ2 / (4n2)
Substituting the known values:
d1 = (4.8 X 10^-7 m) / (4 X 1.25)
Simplifying the expression:
d1 = 9.6 X 10^-8 m
Therefore, the minimum thickness d1 of the film is 9.6 X 10^-8 meter.
(d) To calculate the minimum nonzero thickness d2 of the film such that the resultant intensity of the light reflected back into the air is a maximum, we need to consider the conditions for constructive interference.
Constructive interference occurs when the path difference between the reflected and transmitted rays is equal to an integer multiple of the wavelength in the film (λ2).
The path difference can be calculated as:
Δx = 2n2d2
where Δx is the path difference, n2 is the refractive index of the thin film, and d2 is the thickness of the film.
Since we want to maximize the reflected intensity, we want the path difference to be equal to an integer multiple of λ2.
Setting up the equation:
λ2 = 2n2d2
Solving for d2:
d2 = λ2 / (2n2)
Substituting the known values:
d2 = (4.8 X 10^-7 m) / (2 X 1.25)
Simplifying the expression:
d2 = 1.92 X 10^-7 m
Therefore, the minimum nonzero thickness d2 of the film is 1.92 X 10^-7 meter.