Question

White light is shone on a very thin layer of mica \((n=1.57),\) and above the mica layer, interference maxima for two wavelengths (and no other in between) are seen: one blue wavelength of \(480 \mathrm{nm},\) and one yellow wavelength of \(560 \mathrm{nm} .\) What is the thickness of the mica layer?

Short answer

Answer: The approximate thickness of the mica layer is 967.36 nm.

Step-by-step solution

Recall the thin film interference formula

To find the thickness of the mica layer, we'll use the thin film interference formula that involves the order of interference (m), wavelength of light (λ), refractive index of the thin film (n), and the thickness of the film (t). The formula is given as: m × λ = 2 × n × t × cos(θ) Here, θ is the angle of incidence, but since we are not given any information about it, we'll assume that the incidence angle is small and normal (θ = 0), in which case cos(θ) = 1. Now, the formula becomes: m × λ = 2 × n × t

Plug in the values for the blue wavelength

For the interference maximum for the blue wavelength (λ = 480 nm), we'll plug in the values into the formula: m × 480 nm = 2 × 1.57 × t

Plug in the values for the yellow wavelength

Similarly, for the interference maximum for the yellow wavelength (λ = 560 nm), we'll plug in the values into the formula: m × 560 nm = 2 × 1.57 × t

Determine the common factor

We can notice that the two equations have a common factor, which is the thickness (t) multiplied by a constant value (2 × 1.57). We can write it as follows: m1 × 480 nm = m2 × 560 nm Since there are no other wavelengths in between these maxima, the integers m1 and m2 will likely have a difference of 1. Let's find the integer values.

Find the integer values

Divide the two equations to cancel the thickness (t) and constant value (2 × 1.57): m1 / m2 = 560 nm / 480 nm Therefore, m1 / m2 ≈ 1.167 One possible solution for this ratio is m1 = 7 and m2 = 6.

Calculate the thickness for the blue wavelength

We'll plug in the values for the blue wavelength to find the thickness of the mica layer: 7 × 480 nm = 2 × 1.57 × t t ≈ 967.36 nm

Verify the thickness for the yellow wavelength

We'll plug in the values for the yellow wavelength to verify if it's consistent with the thickness we calculated: 6 × 560 nm = 2 × 1.57 × t t ≈ 967.36 nm Our calculations are consistent.

Conclusion

The thickness of the mica layer is approximately 967.36 nm.

Key concepts

Interference Maxima

When light waves overlap, they can interfere with each other, creating patterns of bright and dark spots known as interference patterns. An interference maximum, or bright fringe, occurs when the waves from different paths constructively interfere, meaning their crests and troughs align to enhance their amplitudes. The result is a bright spot of light.

For thin film interference, such as that seen with mica, the maximum is seen when the optical path difference between the reflected light rays equates to a whole number of wavelengths, ensuring constructive interference. This can be mathematically expressed with the formula pertaining to thin films where the extra phase change upon reflection from the film has been considered. In the case of the mica layer problem, the light waves reflected from the top and bottom surfaces interfere, and certain wavelengths will be amplified (maxima).

Wavelength of Light

The wavelength of light is a fundamental characteristic of electromagnetic waves, representing the distance between two consecutive crests or troughs in a wave pattern. It's typically measured in nanometers (nm), especially for visible light which ranges from approximately 400 nm to 700 nm.

The wavelength determines the color of light, with shorter wavelengths corresponding to blue end of the spectrum and longer wavelengths to the red end. In the case of thin film interference, specific wavelengths are seen as bright fringes due to constructive interference. By knowing the wavelength and the sequence of bright spots seen, one can calculate physical attributes of the film, like its thickness.

Refractive Index

The refractive index of a material, denoted by 'n', is a dimensionless number that describes how light propagates through that medium. It is the ratio of the speed of light in a vacuum to the speed of light in the medium. Materials with a higher refractive index slow down light more and bend light rays to a greater extent upon entering the material at an angle.

In thin film interference, the refractive index of the film is crucial because it affects the optical path length that light travels within the film before reflecting. This optical path length is the physical path multiplied by the refractive index, and it's part of determining where the interference maxima and minima occur, as seen in the formula used to solve for the mica layer's thickness.

Optical Path Difference

The optical path difference (OPD) refers to the difference in the distance travelled by two light waves when they interfere with each other. OPD is not just about the geometric path but also takes the medium's refractive index into account. For example, when light traverses through a medium other than the vacuum, its effective path length is altered by the refractive index.

In the context of thin film interference, the OPD can lead to constructive interference (interference maxima) when it's an integer multiple of the wavelength, and to destructive interference (interference minima) otherwise. The calculation in our exercise hinges on finding the OPD that accounts for the mica layer's thickness and its refractive index. This concept helps us understand why certain colors appear at a particular thickness and why other colors may be absent due to destructive interference.