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Light of wavelength \(\lambda\) strikes a screen containing two slits a distance \(d\) apart at an angle \(\theta_{i}\) to the normal. Determine the angle \(\theta_{m}\) at which the \(m^{\text {th }}\) -order maximum occurs.

Short Answer

Expert verified
\( \theta_{m} = \sin^{-1}(\sin \theta_{i} + \frac{m \lambda}{d}) \)

Step by step solution

01

Understand the Problem

We are dealing with a double-slit interference problem where we need to find the angle \( \theta_{m} \) for the \( m^{\text{th}} \)-order maximum. The light strikes the screen at a certain angle \( \theta_{i} \), and the distance between the slits is \( d \). We need to apply the condition for maxima in double-slit interference.
02

Recall the Condition for Maximum

The condition for constructive interference (maximum intensity) in a double-slit experiment is given by: \[ d \sin \theta_{m} = m \lambda \]where \( d \) is the distance between the slits, \( \theta_{m} \) is the angle for the \( m^{\text{th}} \) maximum, \( m \) is the order number, and \( \lambda \) is the wavelength of the light.
03

Consider Incidence Angle

When the light strikes the slits at an angle \( \theta_{i} \) instead of perpendicular, the condition for the maximum becomes modified to account for this angle. The modified equation is:\[ d(\sin \theta_{m} - \sin \theta_{i}) = m \lambda \] This accounts for the difference due to the angle of incidence \( \theta_{i} \).
04

Solve for \( \theta_{m}\)

Rearrange the equation to solve for \( \theta_{m} \):\[ \sin \theta_{m} = \sin \theta_{i} + \frac{m \lambda}{d} \]Thus, knowing \( \theta_{i} \), \( m \), \( \lambda \), and \( d \), you can find \( \sin \theta_{m} \) and use the inverse sine function to find \( \theta_{m} \).
05

Final Calculation and Transposing

Given the values for \( \theta_{i}, m, \lambda, \) and \( d \), calculate \( \sin \theta_{m} \) and then find \( \theta_{m} \) by taking the inverse sine (\( \sin^{-1} \)):\[ \theta_{m} = \sin^{-1}(\sin \theta_{i} + \frac{m \lambda}{d}) \] This gives the angle at which the \( m^{\text{th}} \) maximum occurs.

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Key Concepts

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

Constructive Interference
Constructive interference occurs when two or more waves combine to produce a wave with a larger amplitude. In the context of the double-slit experiment, this happens when the path difference between the waves coming through two slits is an integer multiple of the wavelength, leading to bright fringes or maxima on the screen.
In mathematical terms, for constructive interference, the formula used is\[ d \sin \theta_m = m \lambda \]where:
  • d is the distance between the slits
  • \(\theta_m\) is the angle to the mth-order maximum
  • m is the order number
  • \(\lambda\) is the wavelength of the light
This equation is a fundamental concept in understanding how interference patterns develop. It describes how the alignment of the waves can affect the observed brightness at different angles.
Angle of Incidence
The angle of incidence is the angle at which incoming light strikes a surface or, in this case, the two slits in the interference experiment described. Denoted as \(\theta_i\), it measures the deviation of the incoming light from a perpendicular line (the normal) to the slits.
When light strikes the slits at this angle instead of perpendicularly, it alters the paths and hence the interference condition. This modification is critical because it affects where the bright fringes appear on the screen.
  • The modified condition for the maxima becomes: \[ d(\sin \theta_m - \sin \theta_i) = m \lambda \]
  • This means that apart from considering the distance and wavelength, the angle at which light hits the slits changes the resulting pattern of light on the screen.
The angle of incidence is thus essential to understanding how light behaves in practical scenarios where alignment may not be perfect.
Wavelength of Light
The wavelength of light, typically denoted by \(\lambda\), is the distance between successive peaks of a wave. It is a crucial factor in determining the interference pattern observed in a double-slit experiment.
It affects the position of the maxima (bright fringes) on the screen, as it determines the path difference required for constructive interference.
  • The color of visible light is directly related to its wavelength, with shorter wavelengths corresponding to blue light and longer wavelengths to red light.
  • In calculations, longer wavelengths will result in maxima that are spaced further apart than shorter wavelengths.
Understanding the wavelength is key to predicting how different colors of light will interfere and create unique patterns.
Order Number
The order number, represented by \(m\), indicates the position of the maxima relative to the central maximum in an interference pattern. It is an integer that labels the sequence of bright fringes.
  • The central maximum is assigned m = 0, the first maxima on either side are m = 1, and so on.
  • The formula \[ d \sin \theta_m = m \lambda \] means that higher-order maxima correspond to larger angles \(\theta_m\).
As the order number increases, the brightness at the maxima weakens because the waves have traveled longer paths and may lose intensity. The order number is crucial for specifying exactly which maximum or minimum we're discussing in a pattern.
Interference Pattern
An interference pattern refers to the regular arrangement of bright and dark bands on a screen resulting from the overlap (interference) of waves from the two slits. This pattern is a direct visual representation of constructive and destructive interference.
  • Constructive interference leads to bright fringes.
  • Destructive interference results in dark fringes where the wave amplitudes cancel each other out.
The double-slit setup showcases this beautifully by creating a series of alternating bright and dark bands that depend on parameters such as the wavelength of light, slit spacing, and the angle of incidence.
The fringe pattern is essential for many scientific and practical applications, from understanding the nature of light to designing optical instruments.

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Most popular questions from this chapter

(II) A uniform thin film of alcohol \((n=1.36)\) lies on a flat glass plate \((n=1.56)\). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for \(\lambda=512 \mathrm{nm}\) and a maximum for \(\lambda=635 \mathrm{nm}\). What is the minimum thickness of the film?

(II) A lens appears greenish yellow \((\lambda=570 \mathrm{nm}\) is strongest) when white light reflects from it. What minimum thickness of coating \((n=1.25)\) do you think is used on such a glass \((n=1.52)\) lens, and why?

(II) A parallel beam of light from a He-Ne laser, with a wavelength \(633 \mathrm{nm}\), falls on two very narrow slits \(0.068 \mathrm{~mm}\) apart. How far apart are the fringes in the center of the pattern on a screen \(3.8 \mathrm{~m}\) away?

Calculate the minimum thickness needed for an antireflective coating \((n=1.38)\) applied to a glass lens in order to eliminate \((a)\) blue \((450 \mathrm{nm}),\) or \((b)\) red \((720 \mathrm{nm})\) reflections for light at normal incidence.

In a compact disc (CD), digital information is stored as a sequence of raised surfaces called "pits" and recessed surfaces called "lands." Both pits and lands are highly reflective and are embedded in a thick plastic material with index of refraction \(n=1.55\) (Fig. 34-34). As a 780-nm wavelength (in air) laser scans across the pit-land sequence, the transition between a neighboring pit and land is sensed by monitoring the intensity of reflected laser light from the CD. At the moment when half the width of the laser beam is reflected from the pit and the other half from the land, we want the two reflected halves of the beam to be \(180^{\circ}\) out of phase with each other. What should be the (minimum) height difference \(t\) between a pit and land? [When this light enters a detector, cancellation of the two out-of-phase halves of the beam produces a minimum detector output.]

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