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Chapter 35: Problem 17

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_0\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0\(^\circ\), what is the intensity? (b) What is the path difference for 480-nm light from the two slits at a point where the phase difference is 60.0\(^\circ\)?

Short Answer

Expert verified
(a) \( \frac{3}{4} I_0 \); (b) 40 nm.

Step by step solution

01

Understanding Interference Intensity

In a two-slit interference experiment, the intensity at any point on the screen is related to the phase difference \( \Delta \phi \) between the light waves coming from the two slits. The formula for intensity \( I \) at a point where the phase difference \( \Delta \phi \) is given by \[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \] where \( I_0 \) is the intensity at the central maximum.
02

Calculate Intensity for Phase Difference 60掳

Given \( \Delta \phi = 60.0^\circ \), convert this into radians as most trigonometric calculations are done in radians. \( \Delta \phi = \frac{\pi}{3} \) radians. Substitute \( \Delta \phi \) into the formula: \[ I = I_0 \cos^2\left(\frac{\pi}{6}\right) \] Calculate \( \cos\left(\frac{\pi}{6}\right) \), which equals \( \frac{\sqrt{3}}{2} \). Thus: \[ I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4} \] Hence, \( I = \frac{3}{4}I_0. \)
03

Understanding Path Difference

The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) by the formula \[ \Delta \phi = \frac{2\pi \Delta x}{\lambda} \] where \( \lambda \) is the wavelength of light.
04

Calculate Path Difference for Phase Difference 60掳

Given \( \Delta \phi = \frac{\pi}{3} \) (for 60掳 phase difference) and \( \lambda = 480 \) nm, substitute into the equation \( \Delta \phi = \frac{2\pi \Delta x}{\lambda} \): \[ \frac{\pi}{3} = \frac{2\pi \Delta x}{480} \] Simplify and solve for \( \Delta x \): \[ \Delta x = \frac{480}{2\cdot 3} = 40 \text{ nm} \] Therefore, the path difference is 40 nm.

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Key Concepts

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

Interference Pattern
In a two-slit interference setup, the term "interference pattern" refers to the series of bright and dark bands or spots that appear on a screen. These bands arise when coherent light waves, emanating from two narrow slits, superpose and interfere with each other.
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When waves overlap, they either reinforce each other, called constructive interference, leading to brighter bands, or they cancel each other out, called destructive interference, resulting in darker regions. This beautiful pattern is a physical demonstration of the wave nature of light.
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Key characteristics of the interference pattern include:
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  • The spacing between the bright and dark bands depends on the wavelength of the light and the distance between the slits.

  • The central bright band, also called the central maximum, is the most intense due to maximum constructive interference.

Intensity Calculation
The intensity of light in a two-slit interference pattern changes depending on where you observe it. This intensity, denoted as \( I \), at any point on the screen is linked to how the phase of the two light waves differs at that point.
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To calculate this, we use the formula:\[I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right)\]Here, \( I_0 \) represents the intensity at the peak of the central maximum, where phase difference is zero.
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  • At a phase difference of 60.0\(^\circ\), converting to radians gives \( \frac{\pi}{3} \).

  • Using this, intensity can be computed as \( I = \frac{3}{4}I_0 \).

  • Thus, the intensity at this point is 75% of the central peak intensity.

Phase Difference
In the context of two-slit interference, phase difference \( \Delta \phi \) is the difference in phase between the wavefronts arriving from the two slits at a particular point on the observation screen.
\(\)
This phase difference is an angle typically measured in degrees or radians, where 360 degrees (or \( 2\pi \) radians) represent a full cycle.
\(\)
  • The phase difference determines whether the interference at that point is constructive or destructive.

  • Constructive interference, resulting in maximum intensity, occurs when \( \Delta \phi \) is a multiple of 2蟺.

  • Destructive interference occurs when \( \Delta \phi \) is an odd multiple of \( \pi \).

Path Difference
Path difference, denoted as \( \Delta x \), is the difference in the distance traveled by two waves from the slits to a point on the screen.
\(\)
This concept is crucial as the path difference leads to a phase difference between the waves, thereby influencing the interference pattern.
\(\)
To determine the path difference corresponding to a specific phase difference, we use the relation:\[\Delta \phi = \frac{2\pi \Delta x}{\lambda}\]Where \( \lambda \) is the wavelength of the light used.
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  • For a phase difference of 60.0\(^\circ\) or \( \frac{\pi}{3} \), the path difference would be 40 nm for 480-nm light.

  • This path difference of 40 nm causes the waves to interfere at a different intensity than at the central maximum.

Wavelength
Wavelength, symbolized as \( \lambda \), is the distance over which the wave's shape repeats. It is a fundamental property of waves, including light, and is typically measured in nanometers (nm) for light waves.
\(\)
The wavelength determines the color of the light and the spacing of the interference pattern formed.
\(\)
  • A smaller wavelength results in closer fringes in the interference pattern.

  • In this context, we are using a light wavelength of 480 nm, which is characteristic of blue and violet light, contributing to the distinctive pattern observed.

  • Wavelength also plays a critical role in calculating both path and phase differences.

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Most popular questions from this chapter

Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

A researcher measures the thickness of a layer of benzene (\(n\) = 1.50) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. She finds that light of wavelength 575 nm is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?

White light reflects at normal incidence from the top and bottom surfaces of a glass plate (\(n\) = 1.52). There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 nm. What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 nm?

A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 nm passing through a plastic substrate of refractive index 1.8. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other (Fig. E35.31). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

Jan first uses a Michelson interferometer with the 606-nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502-nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?
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