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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°

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°

Given \( \Delta \phi = \frac{\pi}{3} \) (for 60° 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.
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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 2π.

  • 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.
\(\)
  • 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.
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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

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference \(minima\) at \(\pm35.20^\circ\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm19.46^\circ\) instead. What is the index of refraction of this liquid?

One round face of a 3.25-m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 mm apart, are made in the center of the black face. When laser light of wavelength 632.8 nm shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 mm wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

Coherent light with wavelength 500 nm passes through narrow slits separated by 0.340 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?

Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?

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