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35.29. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

Short Answer

Expert verified
The angle \( \theta \) of the wedge is approximately \( 0.26^\circ \).

Step by step solution

01

Understand the Problem

We are given two glass plates with a strip of paper creating a thin wedge of air. Light of wavelength \( \lambda = 546 \, \text{nm} \) is used, and 15 interference fringes are observed per centimeter. We need to find the angle \( \theta \) of the wedge formed.
02

Use the Fringe Spacing Formula

The spacing \( \Delta y \) between successive interference fringes is given by \( \Delta y = \frac{\lambda}{2 \sin \theta} \). Since we have 15.0 fringes per cm, the distance between two adjacent fringes is \( \frac{1}{15.0} \) cm.
03

Express Fringe Spacing in Terms of Centimeters

Convert 546 nm to cm: \( \lambda = 546 \times 10^{-7} \text{cm} \). From the given 15 fringes per cm, \( \Delta y = \frac{1}{15.0} \, \text{cm} \).
04

Set Up the Equation for the Angle

Combine the previous results into\[ \frac{1}{15.0} = \frac{546 \times 10^{-7}}{2 \sin \theta} \]. Solve this equation for \( \sin \theta \).
05

Solve for the Angle

Rearrange the equation to find \( \sin \theta \):\[ \sin \theta = \frac{546 \times 10^{-7}}{2 \times \frac{1}{15}} \]. Calculate \( \sin \theta \) using this formula.
06

Find the Angle

Finally, calculate \( \theta \) using \( \theta = \sin^{-1}(\text{{calculated value}}) \). Convert the angle to degrees if necessary.

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Key Concepts

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

Interference Fringes
Interference fringes are an interesting phenomenon that occurs when two or more light waves overlap and combine. This results in areas of constructive and destructive interference, creating a pattern of bright and dark bands. In the case of the glass plates illuminated with light from a mercury-vapor lamp, these fringes form due to the thin wedge of air created by a strip of paper.
  • Constructive Interference: Occurs when the crest of one wave aligns with the crest of another, producing a brighter light.
  • Destructive Interference: Happens when the crest of one wave aligns with the trough of another, resulting in a darker area.
The pattern of interference fringes is dependent on factors like the wavelength of the light and the thickness of the medium causing the interference. In this exercise, the number of fringes per centimeter gives a direct clue about the wedge's angle; thus, understanding interference is crucial to solving the problem.
Wedge Angle
The wedge angle is the very small angle between two surfaces, created here by a thin strip of paper placed between two glass plates. This angle determines the spacing of the interference fringes observed, which can be quite small and require precise measurements.
  • The wedge creates a varying thickness of air that the light must travel through, producing a pattern of interference fringes.
  • This angle is subtle, as seen by the presence of multiple fringes per centimeter.
Optics calculations involving wedge angles often use the formula for fringe spacing, which relates the angle to the wavelength of the light and the fringe pattern: \( \Delta y = \frac{\lambda}{2 \sin \theta} \). This equation helps in finding the wedge angle when the number of fringes per unit length is known, as in the problem at hand.
Mercury-Vapor Lamp
Mercury-vapor lamps are a type of gas discharge lamp that emits light by passing electricity through mercury vapor. They are commonly used in laboratories and street lighting due to their efficiency and emission of bright light.
  • Application in Experiments: In optical experiments like the one described, mercury-vapor lamps are valued for their ability to produce light at specific wavelengths, such as the 546 nm light used here.
  • Characteristic Light: The lamp emits a series of lines in its spectrum, including a distinct green line at 546 nm, which is often used in interference experiments.
Using such a controlled and predictable light source is essential when measuring interference patterns, as it ensures accuracy and repeatability of results in scientific studies.
Wavelength of Light
The wavelength of light is a fundamental concept in optics. It represents the distance between two successive peaks of a wave and determines the color of the light. In this problem, the wavelength of 546 nm (nanometers), characteristic of green light, plays a crucial role in interference phenomena.
  • Role in Interference: The wavelength directly affects the spacing of interference fringes: shorter wavelengths produce more closely spaced fringes.
  • Measurement and Conversion: Often, wavelengths are given in nanometers (nm), but may need to be converted to other units (like centimeters) for calculations.
Understanding how wavelength impacts interference helps us solve problems like determining a wedge angle, as observed in this exercise. Knowing the precise wavelength allows us to reliably use equations to find unknowns, such as the angle of the air wedge forming between the glass plates.

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Most popular questions from this chapter

35.55. A source \(S\) of monochrounatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass, and are separated by a horizontal distance \(x\) . Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

35.3. A radio transmitting station operating at a frequency of 120 \(\mathrm{MHz}\) has two identical antennas that radiate in phase. Antenna \(B\) is 9.00 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(P\) between the antennas and along the line connecting them, a horizontal distance \(x\) to the right of antenna \(A\) . For what values of \(x\) will constructive interference occur at point \(P ?\)

35.30. A plate of glass 9.00 \(\mathrm{cm}\) long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 \(\mathrm{mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 \(\mathrm{nm}\) . How many interference fringes are observed per centimeter in the reflected light?

35.2. Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q\) ?

35.25. Points \(A\) and \(B\) are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a \(12.5-\mathrm{MHz}\) signal horizontally. These transmitters are in phase with other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C\) . directly opposite the midpoint of \(A B\) . The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?

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