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In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_0\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_0\) = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Short Answer

Expert verified
The thickness of the glass is approximately 177.14 nm. The longest wavelength for constructive interference is 496 nm.

Step by step solution

01

Understanding Constructive Interference

Constructive interference occurs when the path difference between two reflected waves is an integer multiple of the wavelength. For thin films, the condition is given by \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) is the thickness of the glass, \(m\) is the order of interference, and \(\lambda\) is the wavelength in vacuum.
02

Setting up Equations for Wavelengths

Given two wavelengths \(\lambda_1 = 496\, \text{nm}\) and \(\lambda_2 = 386\, \text{nm}\), they correspond to consecutive orders of interference, \(m\) and \(m+1\). Set up equations: \(2nt = m \times 496\) and \(2nt = (m+1) \times 386\).
03

Solving for Thickness

Subtract the two equations to eliminate \(2nt\): \((m+1) \times 386 - m \times 496 = 2nt - 2nt = 0\). The equation \(386 = \frac{110}{m+1} + 386\) results from manipulation. Solving \(110/(m+1) = 496 - 386 = 110\), we find that \(m+1 = 2\). Consequently, \(m = 1\).
04

Calculate Thickness

Substitute \(m = 1\) into the equation for the thickness: \(2nt = 1 \times 496\). Calculate \(t\) using \(n = 1.40\): \(t = \frac{496}{2 \times 1.40}\), which simplifies to \(t = \frac{496}{2.8} \approx 177.14\, \text{nm}\).
05

Finding the Longest Wavelength for Constructive Interference

For the longest wavelength, \(m = 0\), so \(2nt = 0 \times \lambda + \lambda\). Thus, \(\lambda = 2nt/m\). Using \(t = 177.14\, \text{nm}\), find \(\lambda_{longest}\) by rearranging and substituting in the same thickness: \(\lambda_{longest} = 2 \times 1.40 \times 177.14\). Simplify to find \(\lambda_{longest} = 496\, \text{nm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constructive Interference
Constructive interference is a fascinating concept in physics that occurs when two or more waves superpose to form a resultant wave of greater amplitude.
  • This principle applies to different contexts, such as sound, water, and light waves.
  • In the context of thin films, like the glass covering a gas chamber, constructive interference happens when the path difference between the waves is an integer multiple of the wavelength.
The equation for this in the case of thin films is given by:\[2nt = m\lambda\]Here, \(n\) is the refractive index of the material, \(t\) is the thickness of the film, \(m\) is the order of interference (an integer), and \(\lambda\) is the wavelength of the light in a vacuum.Understanding this allows scientists to measure the thickness of films by observing the light that reflects off their surfaces.
In experiments like this, when light of certain wavelengths reflects off the surfaces resulting in visible patterns, it either reinforces its intensity due to constructive interference or cancels out due to destructive interference. This helps in calculating exact thickness as seen in our glass example.
Refractive Index
The refractive index (n) of a material is a crucial factor in determining how light travels through it.
  • This index is a measure of how much the speed of light is reduced inside the material compared to its speed in a vacuum.
  • For instance, a refractive index of 1.40 means that light travels 1.40 times slower in the glass compared to a vacuum.
The refractive index plays a critical role in phenomena like interference and refraction. In the context of thin films, the refractive index influences the optical path of light within the material. When light enters a material with a different refractive index, its speed and wavelength change. This adjustment affects how waves combine when they emerge from the film. The difference in optical paths between entering and exiting light alters the phase of the waves, which directly impacts interference patterns observed.
For practical purposes, knowing the refractive index and the behavior of light within the material allows precise calculation of properties like film thickness and helps us predict the conditions under which constructive interference occurs.
Wavelength
Wavelength is a key concept in understanding wave phenomena, such as interference in thin films. It is the distance between two consecutive peaks (or troughs) of a wave.
  • The wavelength of light determines its color when perceived by our eyes, with different wavelengths appearing as different colors.
  • For example, red light has a longer wavelength compared to blue light.
In the realm of interference, the wavelength of light plays a pivotal role in which wavelengths result in constructive interference. When light of a specific wavelength reflects off a thin film with its path altered, it may combine with light that has taken a different route. If the differences in the paths traveled by two waves align such that they are multiples of the wavelength, constructive interference occurs leading to increased intensity at that wavelength.
In experiments using thin films, by carefully adjusting which wavelengths are analyzed, such as 496 nm and 386 nm in this case, scientists can discern thicknesses of materials and explore properties of light as it interacts with differing materials.

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Most popular questions from this chapter

A researcher measures the thickness of a layer of benzene (\(n\) = 1.50) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. She finds that light of wavelength 575 nm is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?

Two very narrow slits are spaced 1.80 \(\mu\)m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda\) = 550 nm? (Hint: The angle \(\theta\) in Eq. (35.5) is \(not\) small.)

Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by 0.200 mm, and the interference pattern is observed on a screen 4.00 m from the slits. (a) What is the width (in mm) of the central interference maximum? (b) What is the width of the first-order bright fringe?

Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?

White light reflects at normal incidence from the top and bottom surfaces of a glass plate (\(n\) = 1.52). There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 nm. What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 nm?

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