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Chapter 35: Problem 53

Consider a two-slit interference pattern, for which the intensity distribution is given by Eq. (35.14). Let \(\theta_{m}\) be the angular position of the \(m\) th bright fringe, where the intensity is \(I_{0}\) . Assume that \(\theta_{m}\) is small, so that \(\sin \theta_{m} \cong \theta_{m}\) . Let \(\theta_{m}^{+}\) and \(\theta_{m}-\) be the two angles on either side of \(\theta_{m}\) for which \(I=\frac{1}{2} I_{0}\) . The quantity \(\Delta \theta_{m}=\left|\theta_{m}^{+}-\theta_{m}^{-}\right|\) is the half-width of the \(m\) th fringe. Calculate \(\Delta \theta_{m} .\) How does \(\Delta \theta_{m}\) depend on \(m\) ?

Short Answer

Expert verified
\( \Delta \theta_{m} = \frac{\lambda}{d} \); it is independent of \( m \).

Step by step solution

01

Understand the given intensity distribution

The problem revolves around the intensity distribution of a two-slit interference pattern. The maximum (bright fringe) intensity is given as \( I_0 \). We have small angles \( \theta_m \) so that \( \sin \theta_m \approx \theta_m \).
02

Apply the condition for half intensity

We are given that \( I = \frac{1}{2} I_0 \) at angles \( \theta_m^+ \) and \( \theta_m^- \) on either side of a specific fringe \( \theta_m \). In a typical interference pattern, this can happen because the intensity expression commonly involves a cosine term: \( I = I_0 \cos^2(\delta/2) \). Setting \( I = \frac{1}{2} I_0 \) implies \( \cos^2(\delta/2) = \frac{1}{2} \).
03

Solve for phase difference condition

The condition \( \cos^2(\delta/2) = \frac{1}{2} \) happens when \( \delta = \pi/2 \pm \pi \). The relationship between phase difference \( \delta \) and angle is \( \delta = kd \sin\theta \), where \( d \) is slit separation and \( k = \frac{2\pi}{\lambda} \).
04

Use small angle approximation

Given \( \theta_m^+, \theta_m^- \), derive \( \delta\theta_m \). With \( \theta_m \) small, \( \sin \theta \approx \theta \) simplifies the phase relationship to \( \delta = k d \theta \). Taking the difference, \( \Delta \theta_m = \theta_m^+ - \theta_m^- = \frac{\pi \lambda}{k d} \).
05

Express the dependence on \( m \)

The angular separation for bright fringes corresponds to \( m \lambda = d \sin \theta \). When \( \theta_m \) is small: \( \theta_m \approx m \frac{\lambda}{d} \). Thus, \( \Delta \theta_m = \frac{\lambda}{d} \) remains constant for small angles, and independent of \( m \).

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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 alternating bright and dark bands that appear on a screen when light passes through two closely spaced slits. This pattern is a classic example of wave interference, where overlapping light waves either constructively or destructively interfere with each other.
  • Bright Fringes: Points where waves overlap, enhancing each other's light, creating bands of bright intensity. This occurs when the path difference between waves is an integer multiple of the wavelength.

  • Dark Fringes: Result from destructive interference, where waves cancel out, leaving regions of reduced intensity or darkness. This happens when the path difference is a half-integer wavelength.

Understanding these fringes helps visualize how light behaves when it acts like a wave, creating patterns that reveal the inherent characteristics of wave interference.
Intensity Distribution
In the context of a two-slit interference pattern, intensity distribution describes how the brightness of the pattern varies across the screen. This distribution is critical in explaining why we see alternate bright and dark bands.
  • Mathematical Expression: For a two-slit interference setup, the intensity at any point is given by the formula: \( I = I_0 \cos^2\left(\frac{\delta}{2}\right) \).

  • This formula suggests that the intensity depends on the phase difference \( \delta \), with \( I_0 \) representing the maximum intensity observed at bright fringes.

  • Half-Intensity Points: The points where the intensity is half of \( I_0 \) occur when the phase difference modifies the path length such that \( \cos^2\left(\frac{\delta}{2}\right) = \frac{1}{2} \), creating the observed pattern on the screen.

By understanding the intensity distribution, one can predict where bright and dim regions will form during a two-slit experiment.
Phase Difference
Phase difference is a crucial concept when studying interference patterns, as it arises from differences in the path lengths traveled by light waves. It determines the type of interference鈥攃onstructive or destructive鈥攖hus affecting the visibility of interference fringes.
  • The phase difference \( \delta \) is given by the relationship \( \delta = kd \sin\theta \), where \( k \) is the wave number \( \frac{2\pi}{\lambda} \), \( d \) is slit separation, and \( \theta \) is the angle of observation.

  • Constructive Interference: Occurs when the phase difference is an integer multiple of \( 2\pi \), leading to bright fringes.

  • Destructive Interference: Occurs when the phase difference is an odd multiple of \( \pi \), resulting in dark fringes.

The phase difference is pivotal because it correlates directly with the optical path difference, influencing the interference pattern's structure and spacing.
Small Angle Approximation
The small angle approximation is a useful mathematical simplification often applied in two-slit interference experiments when the observed angles are relatively small.
  • In this context, the approximation assumes that \( \sin\theta \approx \theta \) for small \( \theta \) when measured in radians.

  • Applications: This simplification is beneficial for calculating positions of interference fringes on a screen, making computations more straightforward by reducing the complexity of trigonometric expressions.

  • Deriving Fringe Width: Using the small angle approximation, the angular fringe width \( \Delta\theta_m \) can be derived as \( \frac{\lambda}{d} \), highlighting that fringe width remains constant and is independent of \( m \) for small angles.

This approximation streamlines analysis, particularly beneficial in academic and practical applications, by allowing easier predictions of interference patterns.

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Most popular questions from this chapter

A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wavelength 525 nm that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 \(\mathrm{nm}\) per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) (red) and 470 nm (blue), passes through two narrow slits separated by \(0.300 \mathrm{mm},\) and the interference pattern is observed on a screen 5.00 \(\mathrm{m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

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.)

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 angle is \(60.0^{\circ} ?\)
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