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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?

Short Answer

Expert verified
The thickness of the glass plate is approximately 156.9 nm.

Step by step solution

01

Identify Key Information and Formula

For constructive interference in a thin film, the condition is \(2nt = m\lambda\) for the wavelengths that produce constructive interference. Here, we know the refractive index \(n = 1.52\), and two wavelengths: 477.0 nm and 540.6 nm. Our task is to find the thickness \(t\) of the plate.
02

Establish Constructive Interference Relationship

The film experiences constructive interference for successive maxima when the order \(m\) increases by one for the next wavelength. Therefore, the equation for the first wavelength is \(2nt = m\lambda_1\) and for the second wavelength \(2nt = (m+1)\lambda_2\).
03

Set Up Equations for the Wavelengths

Substitute the given data into the equations: \(2 \times 1.52 \times t = m \times 477.0\) and \(2 \times 1.52 \times t = (m+1) \times 540.6\).
04

Eliminate Common Factor

Since both equations are equal to \(2nt\), equate them: \(m \times 477.0 = (m+1) \times 540.6\). This allows us to solve for \(m\).
05

Solve for Order \(m\)

Rearrange the equation: \(477m = 540.6m + 540.6\). Simplify to find the order \(m\): \(477 - 540.6 = 540.6m - 477m\). Solve for \(m\): \(63.6 = 63.6m\), so \(m = 1\).
06

Find Thickness of the Glass Plate

Substitute \(m = 1\) into one of the original equations to find \(t\). Using \(2 \times 1.52 \times t = 1 \times 477.0\): solve for \(t\), \(t = \frac{477.0}{2 \times 1.52} = 156.897\) nm.

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Key Concepts

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

Refractive Index
The refractive index, often denoted as "n", is a fundamental concept in optics. It measures how much a medium, such as glass or air, reduces the speed of light compared to its speed in a vacuum. Simply put, it tells us how "bent" or slowed down light becomes when it enters a material.
The refractive index of a medium can be calculated using the formula:
  • \[ n = \frac{c}{v} \]
where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.

In the context of thin film interference, the refractive index helps determine how light behaves as it passes through a film like our glass plate in the exercise above. With a refractive index of 1.52 for glass:
  • Light travels slower in the glass than in the air.
  • It also undergoes bending and reflection at the interfaces of different materials.
  • The refractive index directly impacts the condition for constructive interference.
Wavelength
Wavelength is a key parameter in understanding light waves. Wavelength refers to the distance between successive peaks (or troughs) of the wave. It is usually denoted by the symbol \( \lambda \) and measured in nanometers (nm) when dealing with light.

Light is part of the electromagnetic spectrum, and its wavelength determines its color in the visible spectrum. In our problem, two specific wavelengths are significant: 477.0 nm and 540.6 nm, representing different colors of light.
  • A shorter wavelength means higher frequency and energy.
  • The change in the wavelength when light enters a different medium is due to the refractive index.
  • Knowing the wavelength enables us to calculate when constructive interference occurs within a thin film.
In calculations, the wavelength is essential in determining thin film interference as it affects the wave phase, as it reflects and refracts inside the film.
Constructive Interference
Constructive interference is a phenomenon that occurs when two or more light waves overlap in such a way that they reinforce one another, producing a brighter intensity of light. It is crucial in the study of thin film interference because it determines the conditions when certain colors of light are enhanced.

For thin films, constructive interference happens based on the path difference between the light waves reflected from the top and bottom surfaces of a film. The condition for constructive interference is given by:
  • \[ 2nt = m\lambda \]
where \( t \) is the thickness of the film, \( n \) is the refractive index, \( \lambda \) is the wavelength in air, and \( m \) is the order of interference (an integer).

  • The "m" value increases by one for each successive constructive interference maxima.
  • By analyzing different wavelengths for which constructive interference occurs, we can determine the thickness of the film (as shown in the exercise).
  • This concept is critical for applications like coatings on lenses and anti-reflective coatings.
Overall, understanding constructive interference allows us to predict and manipulate the appearance of colors in films.

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Most popular questions from this chapter

Jan first uses a Michelson interferometer with the 606-nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502-nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

Young's experiment is performed with light from excited helium atoms (\(\lambda\) = 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.300 mm. Their interference pattern is observed on a screen 4.00 m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

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