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Chapter 35: Problem 27

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 of the wedge is approximately 4.095 x 10^-4 radians.

Step by step solution

01

Identify the Problem

We are tasked with finding the angle of a wedge created by two glass plates with a paper strip causing a wedge of air. Given the 546 nm wavelength of light and 15 fringes per centimeter, we must determine the angle of the wedge in radians.
02

Define the Relationship

Interference fringes are formed due to the thin film effect. The number of fringes per unit length is related to the angle by the formula: \( \frac{1}{d} = \frac{2}{\lambda\theta} \), where \( \theta \) is the angle in radians, \( \lambda \) is the wavelength of light used (in meters), and \( d \) is the separation between fringes per unit length (in meters).
03

Convert Units

The wavelength \( \lambda = 546 \text{ nm} = 546 \times 10^{-9} \text{ m} \). The number of fringes per centimeter is 15 fringes/cm, which means \( d = \frac{1}{15 \times 100} \text{ m} \) because 1 cm = 0.01 m. Calculate \( d \).
04

Substitute Values

Using the relationship from Step 2, we substitute the values found in Step 3 into the equation:\\[ \theta = \frac{\lambda}{2d} = \frac{546 \times 10^{-9}}{2 \times \left(\frac{1}{1500}\right)} \].
05

Calculate the Angle

Perform the calculation: \[ \theta = \frac{546 \times 10^{-9} \times 1500}{2} \]. Simplify the expression to find \( \theta \).
06

Finalize and Verify

The calculated value gives the angle of the wedge. Ensure that units are correctly adjusted and check if the result is reasonable given the thin film interference pattern.

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Key Concepts

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

Wedge Angle
When two glass plates have a tiny spacer like a thin strip of paper at one edge, a wedge of air is created between them. The angle of this wedge is referred to as the "wedge angle." This angle, though incredibly small, causes a fascinating phenomenon called thin film interference. Checking the wedge angle involves:
  • Understanding that it's a measure of the slight tilt created by the spacer.
  • Recognizing that different positions along the length of the wedge experience slightly different film thicknesses.
These variations can impact the constructive and destructive interference patterns seen. To compute the wedge angle precisely in such experiments, we often rely on the visible interference fringes formed when light hits the thin wedge.
Interference Fringes
Interference fringes are patterns of dark and light bands created when light waves overlap, either strengthening or canceling each other. This happens when a wedge of air is introduced between the glass plates. You'll see alternating bright and dark lines due to varying film thicknesses:
  • Bright fringes mean constructive interference, where waves add up to produce a stronger light intensity.
  • Dark fringes are the result of destructive interference, where wave peaks meet troughs, canceling each other out.
The number of these fringes per unit length helps determine the wedge angle. Observations often use monochromatic light, like from a mercury-vapor lamp, to ensure clear and specific fringe patterns. The fringes not only help measure the wedge angle but are also vital in various metrology applications, enabling precision measurement in science and engineering.
Optical Path Difference
In thin film interference, the concept of optical path difference (OPD) is pivotal. It denotes the difference in the path lengths traveled by light waves due to varying thicknesses of the film (in this case, the air wedge between glass plates).
  • OPD determines whether interference is constructive or destructive.
  • It is influenced by the wavelength of the light used and the angle of incidence.
When light strikes the wedge, its wavefront is split into two separate paths鈥攐ne reflects off the top surface, while the other travels through the air wedge, reflects off the bottom surface, and then exits back up. Depending on the OPD, these two waves can either reinforce or cancel each other out, determining the interference fringe pattern observed. Thus, understanding OPD is essential for calculating precise wedge angles and effectively interpreting interference phenomena.

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Most popular questions from this chapter

Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) 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 \(\mathrm{nm} ?\)

Two slits spaced 0.260 \(\mathrm{mm}\) apart are placed 0.700 \(\mathrm{m}\) from a screen and illuminated by coherent light with a wavelength of 660 nm. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

Points \(A\) and \(B\) are 56.0 \(\mathrm{m}\) apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5 -MHz signal horizontally. These transmitters are in phase with each 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 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?

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) this film is just thick enough for light with wavelength 582.4 \(\mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength 588.5 \(\mathrm{nm}\) . What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

GPS Transmission. The GPS (Global Positioning System) satellites are approximately 5.18 \(\mathrm{m}\) across and transmit two low-power signals, one of which is at 1575.42 \(\mathrm{MHz}\) (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42 - MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line that extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 \(\mathrm{W} / \mathrm{m}^{2}\) . (a) At how many other angles in the range \(0^{\circ}<\theta<90^{\circ}\) is the intensity also 2.00 \(\mathrm{W} / \mathrm{m}^{2} ?\) (b) Find the four smallest angles in the range \(0^{\circ}<\theta<90^{\circ}\) for which the intensity is 2.00 \(\mathrm{W} / \mathrm{m}^{2} .\) (c) What is the intensity at a point on the circle at an angle of \(4.65^{\circ}\) from the centerline?
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