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Chapter 40: Problem 21

Two flat glass plates are pressed together at the top edge and separated at the bottom edge by a strip of tinfoil. The air wedge is examined in yellow sodium light \((589 \mathrm{~nm})\) reflected normally from its two surfaces. and 42 dark interference fringes are observed. Compute the thickness of the tinfoil.

Short Answer

Expert verified
The thickness of the tinfoil is approximately 12.23 碌m.

Step by step solution

01

Understand the Concept of Interference Fringes

Interference fringes are created when light waves overlap and combine, causing alternating dark and light bands. In this scenario, dark fringes occur when light waves from the top and bottom surfaces cancel each other out.
02

Determine the Formula for Dark Fringes

The condition for dark fringes in an air wedge is given by \(2t = (m + \frac{1}{2}) \lambda\), where \(t\) is the thickness of the air wedge, \(\lambda\) is the wavelength of light, and \(m\) is the fringe order number, starting from 0.
03

Identify Given and Required Values

You know the wavelength \(\lambda = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}\) and that 42 dark fringes are observed, so \(m = 41\). The required value is the thickness \(t\) of the tinfoil.
04

Rearrange Formula to Solve for Thickness

Rearrange the formula to find \(t\): \[t = \frac{(m + \frac{1}{2}) \lambda}{2}\]
05

Calculate the Thickness

Substitute into the formula: \[t = \frac{(41 + \frac{1}{2}) \times 589 \times 10^{-9}}{2}\]. Simplify this to find \(t\) before final calculation, \[t = \frac{41.5 \times 589 \times 10^{-9}}{2} = 1.223065 \times 10^{-5} \text{ m}\].
06

Convert to More Manageable Units (Optional)

Convert the thickness from meters to micrometers (1 m = 1,000,000 碌m), so \(t = 12.23 \text{ 碌m}\).

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Key Concepts

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

Interference Fringes
When we talk about interference fringes, we're diving into one of the captivating phenomena of light behavior. When light waves overlap, they can constructively or destructively combine. This results in a pattern of alternating bright and dark bands known as interference fringes. Here's how it works:
  • **Constructive Interference**: When the crests (tops) of two light waves align, they add up to make a brighter spot.
  • **Destructive Interference**: When a crest meets a trough (bottom) of another wave, they cancel each other out, creating a dark spot.
In our case, we're observing the dark interference fringes created in an air wedge. These occur where the light from the top and bottom glass surfaces of the air wedge cancel out. Interference fringes are crucial in determining thickness because they tell us where this cancellation is happening.
Air Wedge
The air wedge is cleverly simple. Imagine two flat glass plates touching at one edge and separated a bit at the other. This slight separation forms an air wedge.
  • **Creation of Fringes**: As light beams reflect off both surfaces, the differences in the path traveled by each beam lead to interference fringes.
  • **Practical Uses**: Air wedges are used to measure tiny thicknesses and changes, like in the exercise, where they help determine thickness based on the fringe pattern.
The beauty of an air wedge is its simplicity, allowing precise measurements without complex equipment. It鈥檚 a perfect example of how basic principles of optics can provide detailed and valuable information.
Thickness Calculation
In this exercise, we use the interference fringes to calculate the thickness of the tinfoil at the edge of the air wedge.
  • **Formula for Dark Fringes**: The thickness calculation follows the condition for dark fringes: \(2t = (m + \frac{1}{2}) \lambda\), where \(t\) is the air wedge thickness, \(\lambda\) is the wavelength, and \(m\) is the fringe number.
  • **Solving for Thickness**: Rearrange the formula to find \(t\):\[t = \frac{(m + \frac{1}{2}) \lambda}{2}\].
  • **Calculation Steps**: Plug the known values into this formula, with \(\lambda = 589 \times 10^{-9} \text{ m}\) and \(m = 41\), then calculate to attain \(t = 1.223065 \times 10^{-5} \text{ m}\).
These calculations show that even using straightforward principles, we can achieve highly precise measurements, vital for scientific research and technological applications.
Sodium Light Wavelength
Understanding the role of sodium light's wavelength is pivotal in this exercise. Sodium light is often used in interference experiments because of its distinct yellow color and defined wavelength, typically measured at 589 nm.
  • **Why Sodium Light?** This wavelength is stable and specific, which helps achieve highly accurate fringe patterns.
  • **Importance in Calculations**: The precise value of 589 nm (or \(589 \times 10^{-9} \text{ m}\)) is critical when calculating fringe patterns. Any variation in wavelength directly affects the measurement outcome.
  • **Real-World Applications**: Such defined light sources are used in technologies like spectroscopy and optical engineering.
By using a single wavelength like that of sodium light, we can focus on specific patterns and complete accurate measurements, as highlighted in this exercise.

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Most popular questions from this chapter

Green light of wavelength \(500 \mathrm{~nm}\) is incident normally on a grating. and the second-order image is diffracted \(32.0^{\circ}\) from the normal. How many lines \(/ \mathrm{cm}\) are marked on the grating?

Red light of wavelength \(644 \mathrm{~nm}\), from a point source, passes through two parallel and narrow slits which are \(1.00 \mathrm{~mm}\) apart. Determine the distance between the central bright fringe and the third dark interference fringe formed on a screen parallel to the plane of the slits and \(1.00 \mathrm{~m}\) away.

A mixture of yellow light of wavelength \(580 \mathrm{~nm}\) and blue light of wavelength \(450 \mathrm{~nm}\) is incident normally on an air film \(290 \mathrm{~nm}\) thick. What is the color of the reflected light?

Monochromatic light from a point source illuminates two narrow, horizontal, parallel slits. The centers of the two slits are \(a=0.80 \mathrm{~mm}\) apart, as shown in Fig. \(40-3\). An interference pattern forms on the screen, \(50 \mathrm{~cm}\) away. In the pattern, the bright and dark fringes are evenly spaced. The distance \(y_{1}\) shown is \(0.304 \mathrm{~mm}\). Compute the wavelength \(\lambda\) of the light. Notice first that Fig. \(40-3\) is not to scale. The rays from the slits would actually be nearly parallel. We can therefore use the result of Problem \(40.2\) with \(\left(r_{1}-r_{2}\right)=m \lambda\) at the maxima (bright spots), where \(m=0, \pm 1, \pm 2, \ldots\) Then $$\sin \theta=\frac{\left(r_{1}-r_{2}\right)}{a} \quad \text { becomes } \quad m \lambda=a \sin \theta_{m}$$ Or, alternatively, we could use the grating equation, since a double slit is simply a grating with two lines. Both approaches result in \(m \lambda=a \sin \theta_{m}\) We know that the distance from the central maximum to the first maximum on either side is \(0.304 \mathrm{~mm}\). Therefore, from Fig. \(40-3\). $$\sin \theta_{1}=\frac{0.0304 \mathrm{~cm}}{50 \mathrm{~cm}}=0.000608$$ Then, for \(m=1\), $$m \lambda=a \sin \theta_{m} \quad \text { becomes } \quad(1) \lambda=\left(0.80 \times 10^{-3} \mathrm{~m}\right)\left(6.08 \times 10^{-4}\right)$$ from which \(\lambda=486 \mathrm{~nm}\), or to two significant figures, \(0.49 \times 10^{3} \mathrm{~nm}\).

A single slit of width \(0.140 \mathrm{~mm}\) is illuminated by monochromatic light, and diffraction bands are observed on a screen \(2.00 \mathrm{~m}\) away. If the second dark band is \(16.0 \mathrm{~mm}\) from the central bright band, what is the wavelength of the light?
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