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What is the minimum (non-zero) thickness for the air layer between two flat glass surfaces if the glass is to appear dark when 680 -nm light is incident normally? What if the glass is to appear bright?

Short Answer

Expert verified
Minimum non-zero thickness for dark is 170 nm; for bright is 340 nm.

Step by step solution

01

Understanding the Phenomenon

For the glass to appear dark, destructive interference must occur. This means the path difference between the reflected beams from the top and bottom surfaces of the air layer must be a half-integer multiple of the wavelength.
02

Calculating Minimum Thickness for Destructive Interference

The condition for destructive interference in a thin film is given by the formula: \(2t = (m + \frac{1}{2})\lambda/n\), where \(t\) is the thickness, \(\lambda\) is the wavelength, \(n\) is the refractive index of air (approximately 1), and \(m\) is an integer starting from 0. For the minimum non-zero thickness, let \(m = 0\). So, \(2t = (0 + \frac{1}{2}) \times 680\, nm\). Solving for \(t\), we get \(t = \frac{680}{4} = 170\, nm\).
03

Calculating Minimum Thickness for Constructive Interference

For the glass to appear bright, constructive interference must happen. The condition for constructive interference is \(2t = m\lambda/n\). For the smallest non-zero \(t\), we choose \(m = 1\). Solving \(2t = 1 \times 680\) gives \(t = \frac{680}{2} = 340\, nm\).
04

Final Verification

Recheck calculated values to ensure understanding: Destructive interference occurs at \(170\, nm\) and constructive at \(340\, nm\) for normal incidence where the wavelength is unaffected by air.

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Key Concepts

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

Destructive Interference
Destructive interference occurs when two waves meet and their amplitudes cancel each other out, leading to a reduction in light intensity. In the context of thin film interference, this happens when the path difference between the light waves reflecting off the two surfaces of a film, such as the air layer between glass surfaces, is a half-integer multiple of the wavelength of the light being used.

This path difference causes the peaks of one wave to align with the troughs of the other, resulting in cancelation. The condition for destructive interference in a thin film is described by the formula: \[2t = (m + \frac{1}{2})\frac{\lambda}{n}\]where:
  • \( t \) is the thickness of the film
  • \( \lambda \) is the wavelength of the incident light
  • \( n \) is the refractive index of the medium (approximately 1 for air)
  • \( m \) is an integer (starting from 0 for the minimum nonzero thickness)
To achieve destructive interference and consequently cause the glass to appear dark, the minimum non-zero thickness of the air layer can be calculated by setting \( m = 0 \), resulting in a thickness of \( 170 \, nm \) for a wavelength of \( 680 \, nm \).
Constructive Interference
Constructive interference is the opposite of destructive interference. It occurs when two waves meet in phase, meaning their peaks and troughs align perfectly, leading to an increase in amplitude and, hence, light intensity. In thin film interference, this results in the surfaces appearing bright.

For constructive interference to take place, the path difference between the reflected waves must be a whole number of wavelengths. This condition in thin films is expressed by:\[2t = m\frac{\lambda}{n}\]where:
  • \( t \) is the thickness of the thin film
  • \( \lambda \) is the wavelength of light
  • \( n \) is the refractive index of the film (\( n \approx 1 \) for air)
  • \( m \) is an integer (starting from 1 for the smallest nonzero thickness)
By choosing \( m = 1 \), you achieve the minimum non-zero thickness needed for constructive interference. This setting reveals a thickness of \( 340 \, nm \) for light with a wavelength of \( 680 \, nm \). This precise alignment causes the reflected light to reinforce each other, leading to a bright appearance on the glass surfaces.
Path Difference
Path difference is a critical concept in understanding thin film interference. It refers to the difference in the paths traveled by two rays of light as they reflect off different surfaces of a film. This difference determines whether the interference will be constructive or destructive.

When light encounters a thin film, a portion of it reflects off the upper surface, and part travels through the film, reflects off the lower surface, and exits back into the original medium. The path difference is essentially how much longer one path is compared to the other, influenced largely by the thickness of the film and the angle of incidence.

If the path difference leads to a full wavelength (or multiple full wavelengths), the waves reinforce each other, resulting in constructive interference. However, if the path difference is a half-wavelength (or an odd multiple of half-wavelengths), destructive interference occurs, because the light waves are out of phase and cancel each other out.
  • Path difference plays a pivotal role in applications requiring precise control of light reflection, such as anti-reflective coatings and sensors.
  • By adjusting the thickness of the film or changing the angle of incidence, one can manipulate the path difference to achieve the desired form of interference.

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Most popular questions from this chapter

(II) Light of wavelength \(680 \mathrm{nm}\) falls on two slits and produces an interference pattern in which the third-order bright fringe is \(38 \mathrm{~mm}\) from the central fringe on a screen \(2.6 \mathrm{~m}\) away. What is the separation of the two slits?

(1I) A total of 31 bright and 31 dark Newton's rings (not counting the dark spot at the center) are observed when 560 -nm light falls normally on a planoconvex lens resting on a flat glass surface (Fig. 18). How much thicker is the center than the edges?

(II) A thin oil slick \(\left(n_{\mathrm{o}}=1.50\right)\) floats on water \(\left(n_{\mathrm{w}}=1.33\right) .\) When a beam of white light strikes this film at normal incidence from air, the only enhanced reflected colors are red \((650 \mathrm{nm})\) and violet \((390 \mathrm{nm}) .\) From this information, deduce the (minimum) thickness \(t\) of the oil slick.

Light of wavelength \(\lambda\) strikes a screen containing two slits a distance \(d\) apart at an angle \(\theta_{i}\) to the normal. Determine the angle \(\theta_{m}\) at which the \(m^{\text {th }}\) -order maximum occurs.

(II) A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright fringes spaced \(5.00 \mathrm{~mm}\) apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are \(5.14 \mathrm{~mm}\) apart. What is the wavelength of light produced by the pointer?

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