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Chapter 35: Problem 21

Coherent light with wavelength 500 nm passes through narrow slits separated by 0.340 \(\mathrm{mm} .\) At a distance from the slits large compared to their separation, what is the phase difference (in radians) in the light from the two slits at an angle of \(23.0^{\circ}\) from the centerline?

Short Answer

Expert verified
The phase difference is approximately 1.671 radians.

Step by step solution

01

Identify Given Parameters

The given parameters are:- Wavelength, \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \)- Slit separation, \( d = 0.340 \text{ mm} = 0.340 \times 10^{-3} \text{ m} \)- Angle, \( \theta = 23.0^\circ \)
02

Convert Angle to Radians

We need to convert the angle from degrees to radians using the conversion: \( 1^\circ = \frac{\pi}{180} \text{ radians} \).\[\theta = 23.0^\circ = 23.0 \times \frac{\pi}{180} \approx 0.4014 \text{ radians}\]
03

Calculate Path Difference

The path difference \( \Delta x \) for light from two slits at angle \( \theta \) is given by:\[\Delta x = d \sin \theta\]Substitute the values:\[\Delta x = 0.340 \times 10^{-3} \times \sin(0.4014) \approx 0.340 \times 10^{-3} \times 0.3907 \approx 1.329 \times 10^{-4} \text{ m}\]
04

Calculate Phase Difference

The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) and wavelength \( \lambda \) by:\[\Delta \phi = \frac{2 \pi}{\lambda} \times \Delta x\]Substitute the path difference and wavelength:\[\Delta \phi = \frac{2 \pi}{500 \times 10^{-9}} \times 1.329 \times 10^{-4}\]\[\Delta \phi \approx \frac{6.2832}{500 \times 10^{-9}} \times 1.329 \times 10^{-4} \approx 1.671 \text{ radians}\]

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Key Concepts

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

Wavelength
The concept of wavelength is crucial in understanding the behavior of waves, including light waves in a double-slit experiment. Wavelength (\( \lambda \)) is defined as the distance between consecutive crests or troughs of a wave. For instance, in the case of visible light, different wavelengths correspond to different colors.
  • In the exercise, we are dealing with a wavelength of 500 nanometers (nm), which falls into the visible spectrum, specifically around the green light region.
  • This small distance, expressed in terms of nanometers, highlights the precision required in optical experiments.
  • Knowing the wavelength allows us to calculate other vital parameters such as path difference and phase difference.
Wavelength is a fundamental property that influences how the wave behaves as it passes through various media or interacts with surfaces, like slits in the double-slit experiment.
Phase Difference
Phase difference (\( \Delta \phi \)) is an essential concept when analyzing wave interference, especially in a double-slit setup. It refers to the difference in phase between two waves arriving at a point, which occurs when waves travel different paths.
  • In our scenario, we calculate phase difference to determine how out of step the waves are after passing through the slits.
  • It's quantified using radians, where a full wave cycle is represented as \( 2\pi \) radians.
  • The phase difference determines if the waves interfere constructively or destructively at a particular point.
Constructive interference (waves in phase) produces bright bands on a screen, whereas destructive interference (waves out of phase) produces dark bands. Calculating the phase difference is crucial to predict and understand the pattern produced in such experiments.
Path Difference
Path difference (\( \Delta x \)) in a double-slit experiment is the difference in distance traveled by two waves before reaching the same point. This difference directly affects the phase difference and consequently the interference pattern produced.
  • Path difference is calculated using the formula: \[ \Delta x = d \sin \theta \] where \( d \) is the distance between slits and \( \theta \) is the angle of observation.
  • In our exercise, solving for the path difference helps us see how much additional distance one wave travels compared to the other.
  • This measurement is crucial because it directly influences the phase difference, determining whether the combined waves reinforce or cancel each other.
Understanding path difference allows us to predict the positions of maxima and minima in the interference pattern.
Angle of Incidence
In the context of double-slit interference, the angle of incidence (\( \theta \)) is the angle at which the wavefront approaches the slits or the angle at which we observe the interference pattern. It is critical in determining where different phases of light will appear on the detection screen.
  • Angle of incidence is often measured in degrees, but for calculation purposes, especially in physics problems, it's converted to radians.
  • It helps to define the path difference as it affects how much more one path travels compared to another.
  • The angle influences the spread of the interference bands on the observation screen, meaning small angular changes shift the location of bright and dark spots.
Understanding the angle of incidence allows for insight into where the interference effects will be strongest and how they can be modulated or adjusted.
Radians
Radians are a unit of measure for angles and are especially useful in wave physics because they simplify the mathematics involved with periodic functions. One complete rotation around a circle is \( 2\pi \) radians, which equates to 360 degrees.
  • In the exercise, we must convert angles from degrees to radians for our calculations.
  • This is done using the conversion factor: \( 1 \text{ degree} = \frac{\pi}{180} \text{ radians} \).
  • Using radians in calculations involving phase and path differences allows us to directly relate the angular measures to the wave phenomena.
Radians are invaluable in the study of waves because they provide a natural way to handle the math of circles and cycles, making them ideal for understanding periodic motion like light interference.

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Most popular questions from this chapter

A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of TiO \(_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO\(_{2}\) film

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.)

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 \(\mathrm{nm} ?\) The index of refraction of the film is \(1.33,\) and there is air on both sides of the film.

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 \(\mathrm{nm} ?\)

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?
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