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Chapter 26: Problem 13

Two thin parallel slits that are \(0.0116 \mathrm{~mm}\) apart are illuminated by a laser beam of wavelength \(585 \mathrm{nm}\). (a) How many bright fringes are there in the angular range of \(0<\theta<20^{\circ} ?\) (b) How many dark fringes are there in this range?

Short Answer

Expert verified
(a) 7 bright fringes; (b) 6 dark fringes.

Step by step solution

01

Convert Units

Convert the given measurements into consistent units. The slit separation \(d\) is \(0.0116\, \text{mm} = 0.0116 \times 10^{-3}\, \text{m}\). The wavelength \(\lambda\) is \(585\, \text{nm} = 585 \times 10^{-9}\, \text{m}\).
02

Bright Fringe Condition

Use the formula for the position of bright fringes: \(d\sin\theta = m\lambda\) where \(m\) is the fringe order. We solve for \(m\) using \(\theta=20^{\circ}\): \(0.0116 \times 10^{-3} \sin(20^{\circ}) = m \times 585 \times 10^{-9}\).
03

Calculating m for Bright Fringes

Calculate \(\sin(20^{\circ})\approx0.3420\). The equation becomes \(0.0116 \times 10^{-3} \times 0.3420 = m \times 585 \times 10^{-9}\). Solving for \(m\) gives \(m\approx6.79\). Since \(m\) must be an integer, the values range from \(m=0\) to \(m=6\), giving 7 bright fringes.
04

Dark Fringe Condition

Use the formula for the position of dark fringes: \(d\sin\theta = (m+0.5)\lambda\). Solving similarly with \(\theta=20^{\circ}\) gives: \(0.0116 \times 10^{-3} \times 0.3420 = (m+0.5) \times 585 \times 10^{-9}\).
05

Calculating m for Dark Fringes

Rearrange to find \(m\): \(m+0.5 \approx 6.29\). Solving for integer \(m\), it ranges from \(m=0\) to \(m=5\), giving 6 dark fringes.

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Key Concepts

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

Thin Parallel Slits
Thin parallel slits are a fundamental component in experiments investigating wave interference patterns. When a beam of light passes through these slits, it creates alternating patterns of light and dark known as interference patterns. Each slit serves as a source of waves that interfere with one another. This phenomenon can be visualized when monochromatic light, such as a laser beam, is used.
  • Light waves emanate from each slit.
  • These waves superimpose on each other leading to constructive and destructive interference.
  • Constructive interference creates bright fringes, while destructive interference results in dark fringes.
These interference patterns are mainly determined by the slit separation and the wavelength of the incident light. Thin slits help provide clear, distinguishable interference patterns necessary for understanding wave behavior.
Bright Fringes
Bright fringes occur as a result of constructive interference. When the light waves emerging from thin parallel slits are in phase, they add up to form these areas of increased intensity.
  • The condition for bright fringes is given by: \( d\sin\theta = m\lambda \).
  • Here, \( d \) is the distance between slits and \( \lambda \) is the wavelength.
  • \( m \) represents the order of the fringe, an integer, indicating the sequence of bright bands.
For example, a bright fringe at \( m=0 \) is known as the zeroth-order fringe, directly in line with the source. As \( m \) increases, the fringes spread farther apart. Each order corresponds to a specific angle \( \theta \) in the angular field of the interference pattern.
Dark Fringes
Dark fringes arise due to destructive interference, where light waves from the slits are out of phase. At these points, the waves cancel each other out, leading to areas of zero or minimum intensity.
  • The condition for dark fringes is \( d\sin\theta = (m+0.5)\lambda \).
  • \( m \) is again the order of the fringe, but includes a half-integer shift to indicate the alternate positioning compared to bright fringes.
Dark fringes lie between bright fringes and help visualize the alternating pattern. Understanding these positions helps in calculating interference effects in optical setups like diffraction gratings.
Wavelength Conversion
For the calculations involving interference patterns, it's crucial to convert measurements into consistent units. Wavelength, typically given in nanometers (nm), needs to be converted to meters (m) for uniformity in equations.
  • 1 nanometer equals \( 10^{-9} \) meters.
  • This conversion is essential for aligning the units of distance in the interference formulas.
Such conversions ensure that calculations for angles and fringe positions are precise, especially when using delicate instruments that rely on wave properties.
Fringe Order Calculation
Fringe order calculation involves determining the sequence number \( m \) for both bright and dark fringes. This calculation links directly to the interference conditions and the angular range involved.
  • With \( \theta = 20^{\circ} \), \( \sin\theta \approx 0.3420 \).
  • For bright fringes, solve \( d\sin\theta = m\lambda \) for integer values of \( m \).
  • For dark fringes, use \( d\sin\theta = (m+0.5)\lambda \) and calculate for integer \( m \).
Deciphering \( m \) provides insights into the number of observable fringes within a particular range. It is critical for predicting patterns in practical scenarios, such as quality control in optics and other precision industries.

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Most popular questions from this chapter

The lenses of a particular set of binoculars have a coating with index of refraction \(n=1.38,\) and the glass itself has \(n=1.52 .\) If the lenses reflect a wavelength of \(525 \mathrm{nm}\) the most strongly, what is the minimum thickness of the coating?

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

Even if the lenses of our eyes functioned perfectly, our vision would still be limited due to diffraction of light at the pupil. Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at his near point of \(25.0 \mathrm{~cm}\) with a pupil diameter of 2.00 \(\mathrm{mm} ?\) Assume that the light has a wavelength of \(550 \mathrm{nm}\). (To get a reasonable estimate without having to go through complicated calculations, we'll ignore the effect of the fluid in the eye.) Based on your answer, does it seem that diffraction plays a significant role in limiting our visual acuity?

Interference occurs not only with light waves but also with all frequencies of electromagnetic waves and all other types of waves, such as sound and water waves. Suppose that your physics professor sets up two sound speakers in the front of your classroom and uses an electronic oscillator to produce sound waves of a single frequency. When she turns the oscillator on (take this to be its original setting), you and many students hear a loud tone while other students hear nothing. (The speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\).) The professor then adjusts the apparatus. The frequency that you hear does not change, but the loudness decreases. Now all of your fellow students can hear the loud tone. What did the professor do? A. She turned off the oscillator. B. She turned down the volume of the speakers. C. She changed the phase relationship of the speakers. D. She disconnected one speaker.

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 \(575 \mathrm{nm}\) is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?
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