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Chapter 35: Problem 17

35.17. Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ? )\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Short Answer

Expert verified
There are 39 bright fringes, and the most distant one occurs at 73.6 degrees.

Step by step solution

01

Understanding the Problem

We have two slits with a separation of 0.0116 mm and light with a wavelength of 585 nm. We need to find the total number of bright fringes and the angle of the farthest fringe.
02

Using the Condition for Bright Fringes

The condition for constructive interference (bright fringes) is given by the formula \(d \sin \theta = m \lambda\), where \(d\) is the separation between the slits, \(\lambda\) is the wavelength, and \(m\) is the order of the fringe.
03

Determining Maximum \(\sin\theta\) Value

Since \(\sin \theta\) can have a maximum value of 1, we set \(d \cdot 1 = m \lambda\). Therefore, the largest possible integer value of \(m\), noted as \(m_{\max}\), satisfies \(m \lambda \leq d\).
04

Calculating Maximum Order \(m_{\max}\)

Convert the values: \(d = 0.0116 \ mm = 11.6 \times 10^{-6} \ m\) and \(\lambda = 585 \ nm = 585 \times 10^{-9} \ m\). Solve \(m \leq \frac{d}{\lambda}\) to find \(m_{\max} = \left\lfloor \frac{11.6 \times 10^{-6}}{585 \times 10^{-9}} \right\rfloor\). This evaluates to \(m_{\max} = 19\).
05

Finding Total Number of Bright Fringes

Considering the central fringe (\(m=0\)) and symmetrical fringes on both sides, the total number of bright fringes is \(2 \times m_{\max} + 1 = 2 \times 19 + 1\), which equals 39 fringes.
06

Calculating Angle for the Most Distant Fringe

The most distant fringe corresponds to \(m_{\max} = 19\). Using \(\sin \theta = \frac{m \lambda}{d} = \frac{19 \times 585 \times 10^{-9}}{11.6 \times 10^{-6}}\), solve for \(\theta\). Thus, \(\theta = \arcsin(0.957)\), which evaluates to approximately 73.6 degrees.

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Key Concepts

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

Double Slit Experiment
The double slit experiment is a famous demonstration in physics that showcases the wave nature of light. It involves shining a beam of light, such as from a laser, through two close parallel slits. This setup results in an interference pattern on a screen that is placed some distance away. The distinctive pattern typically displays alternating bright and dark fringes.

Here鈥檚 why this happens:
  • Light waves from the two slits spread out and overlap.
  • When these waves meet, they interfere with each other.
  • This interference can be constructive or destructive, leading to bright and dark fringes.
Clear observation of this experiment underscores a fundamental principle of quantum mechanics: light behaves both like particles and waves. Every time you encounter a bright spot on the screen, it's a result of constructive interference of waves.
Constructive Interference
Constructive interference occurs when two or more wavefronts combine to form a wave of greater amplitude than the individual waves. In the context of the double slit experiment, this means brighter spots on the screen.

This phenomenon occurs when the path difference between the waves from the two slits is a whole number multiple of the light's wavelength. Mathematically, it's expressed as:
  • Condition: \( d \sin \theta = m \lambda \)
  • Here, \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe from the central axis, \( \lambda \) is the wavelength of the light, and \( m \) is the fringe order or an integer.
When these conditions are met, the resulting wave at that point is intensified, leading to a bright fringe. Understanding constructive interference aids in grasping why light from slits creates bright lines periodically.
Fringe Order
Fringe order, denoted as \( m \), is a pivotal concept in understanding the interference pattern of the double slit experiment. Essentially, it refers to the number assigned to a bright or dark fringe based on its position relative to the central maximum. The central fringe is typically \( m = 0 \), with others numbered sequentially on either side.

In the bright fringe formula \( d \sin \theta = m \lambda \), \( m \) determines how far away a particular fringe is from the central bright fringe, which is equivalently displacing it from the central maximum. Positive and negative \( m \) values describe symmetrical fringe locations.
  • Central Fringe: \( m = 0 \)
  • First order fringe: \( m = \pm 1 \)
  • Second order fringe: \( m = \pm 2 \)
Identifying the fringe order helps calculate not only the number of bright fringes but also their respective angles; thus, it is crucial for problems involving interference patterns.

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Most popular questions from this chapter

35.30. A plate of glass 9.00 \(\mathrm{cm}\) long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 \(\mathrm{mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 \(\mathrm{nm}\) . How many interference fringes are observed per centimeter in the reflected light?

35.15. Coherent light with wavelength 600 \(\mathrm{nm}\) passes through two very narrow slits and the interference pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slits. The first-order bright fringe is at 4.84 \(\mathrm{mm}\) 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?

35.52. Red light with wavelength 700 \(\mathrm{nm}\) is passed through a two- slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

35.39. The radius of curvature of the convex surface of a planoconvex lens is 95.2 \(\mathrm{cm}\) . The lens is placed convex side down on a perfectly flat glass plate that is illuminated from above with red light having a wavelength of 580 \(\mathrm{nm}\) . Find the diameter of the second bright ring in the interference pattern.

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