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Light of wavelength 585 \(\mathrm{nm}\) falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calclating all the anglest (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be 7 (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Short Answer

Expert verified
There are 226 dark fringes. The most distant dark fringe occurs at \(\theta \approx 85.7^{\circ}\).

Step by step solution

01

Understanding the Problem

We need to determine the number of dark fringes on a screen when light of a specific wavelength passes through a slit. This involves finding conditions for destructive interference, which occurs at dark fringes.
02

Apply Condition for Destructive Interference

Destructive interference occurs at angles where \(a \sin \theta = m \lambda\), where \(a\) is the slit width (0.0666 mm), \(\lambda\) is the wavelength (585 nm), and \(m\) is an integer representing the order of the dark fringe.
03

Convert Units

Convert all measurements to consistent units. Changing nm to mm: 585 nm = 0.000585 mm.
04

Find Range of Sin(θ)

The sine of an angle ranges from -1 to 1. Since \(\sin \theta\) cannot exceed 1, \(m\) should be such that \(m \lambda/a \leq 1\).
05

Calculate Maximum Order Number m

Substitute the known values into the inequality: \(m \cdot 0.000585 \, \text{mm} / 0.0666 \, \text{mm} \leq 1\). Solve for \(m\) to find the maximum integer value.
06

Solve Inequality for m

Calculate \(m \leq \frac{0.0666}{0.000585} \approx 113.84\). Since \(m\) must be an integer, the maximum value is \(m = 113\).
07

Count Total Dark Fringes

Each value of \(m\) produces a dark fringe on either side of the central maximum, except \(m = 0\) which corresponds to the central bright spot. The total number of dark fringes is 2\(m\) for \(m = 1, 2, ... , 113\). So, the total number of dark fringes is \(2 \times 113 = 226\).
08

Determine the Most Distant Dark Fringe Angle

Using the relation \(m = 113\), find \(\theta\) using \(\sin \theta = \frac{113 \times 0.000585}{0.0666} \approx 0.996\). \(\theta = \arcsin(0.996)\).

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Key Concepts

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

Destructive Interference
Destructive interference is a key concept in wave optics, describing how waves (such as light) can cancel each other out. This occurs when two waves are out of phase, meaning their peaks and troughs are misaligned. For two waves to completely cancel out, a crest of one must meet a trough of the other. In the context of the single slit diffraction problem, destructive interference results in dark fringes on the screen. These are areas where light waves interfere in such a way that their combined amplitude is zero. The condition for destructive interference in a single slit diffraction pattern is given by the equation:
  • \[ a \sin \theta = m \lambda \]
  • where \(a\) is the slit width, \(\lambda\) is the wavelength of the light, and \(m\) is an integer known as the order of the dark fringe.
This equation effectively determines where the dark fringes will occur through the interference of light.
Single Slit Diffraction
Single slit diffraction refers to the spreading of light as it passes through a narrow opening. When light encounters a small slit, it doesn't just continue in a straight line, due to diffraction, it spreads out. This spreading causes a pattern of bright and dark regions on a screen placed at some distance from the slit.

In this pattern:
  • The central bright spot, or maximum, is the widest and brightest because of the constructive interference where waves align in phase.
  • Moving outward, the light intensity decreases creating alternate dark and bright regions. These darker regions are caused by destructive interference.
The width of the slit, along with the wavelength of the light, directly influences this diffraction pattern. The formula \(a \sin \theta = m \lambda\) helps predict the angles at which these dark fringes (destructive interference) occur.
Wave Optics
Wave optics, or physical optics, focuses on understanding the behavior of light as a wave. Unlike ray optics, which treats light as straight rays, wave optics takes into account phenomena such as interference, diffraction, and polarization that arise from the wave nature of light.

Wave optics is crucial for explaining the interference patterns seen in the single slit diffraction experiment.
  • The approach helps clarify how light waves can spread out and overlap, producing distinct patterns of bright and dark regions.
  • It also predicts how changes in parameters like slit width and light wavelength will alter the pattern.
This deeper understanding is essential for applications involving precision and control over light behavior, such as in optical instruments and modern technology.
Diffraction Gratings
Diffraction gratings are optical components with a large number of parallel slits situated very close together. They are used to disperse light into its component wavelengths, creating a spectrum. In the diffraction grating, the principle is similar to single slit diffraction, but vastly amplified.

Instead of one slit, multiple slits cause the waves to interfere, leading to very sharp, bright lines at specific angles. These sharp lines are much more defined than those in a single slit experiment and are used to measure wavelengths precisely.
  • Each order of interference maximum corresponds to a specific angle, which depends on the wavelength of the light and the spacing between the slits.
  • The condition remains similar: \(d \sin \theta = m \lambda\), where \(d\) is the distance between adjacent slits.
With diffraction gratings, employing this condition allows scientists to precisely determine the wavelengths of light present in a sample, making them invaluable tools in spectroscopy and various scientific analyses.

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Most popular questions from this chapter

Light of wavelength 633 \(\mathrm{nm}\) from a distant source is incident on a slit 0.750 \(\mathrm{mm}\) wide, and the resulting diffraction pattern is observed on a screen 3.50 \(\mathrm{m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

Observing Jupiter. You are asked to design a space telescope for earth orbit. When Jupiter is \(5.93 \times 10^{8} \mathrm{km}\) away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 \(\mathrm{km}\) apart. What minimum-diameter mirror is required? Assume a wavelength of 500 \(\mathrm{nm}\) .

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 \(\mathrm{mm}\) wide. At the point in the pattern \(3.25^{\circ}\) from the center of the central maximum, the total phase difference between wavelets from the tup and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_{0} ?\)

Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm}\) . The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is \(36.5 \mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute). (a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 550 -nm light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10- diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope ( 5 -m diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

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