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In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe is \(5.4^{\circ} .\) Find the ratio \(d / \lambda\) of the slit separation \(d\) to the wavelength \(\lambda\) of the light.

Short Answer

Expert verified
The ratio \( \frac{d}{\lambda} \) is approximately 15.96.

Step by step solution

01

Understanding the Problem

In Young's double-slit experiment, the angle to a dark fringe is given. We aim to find the ratio of the slit separation to the wavelength, denoted as \( \frac{d}{\lambda} \). The given angle is \( 5.4^{\circ} \) for the second dark fringe.
02

Identify the Formula

The formula for the angular position of the dark fringes in a double-slit experiment is given by \( d\sin(\theta) = (m+\frac{1}{2})\lambda \), where \( m \) is the order number of the dark fringe. For the second dark fringe, \( m = 1 \). Substituting, we get \( d\sin(\theta) = 1.5\lambda \).
03

Substitute Known Values

Substitute the given angle \( \theta = 5.4^{\circ} \) into the equation: \( d\sin(5.4^{\circ}) = 1.5\lambda \).
04

Calculate \(\sin(\theta)\)

Calculate \( \sin(5.4^{\circ}) \). Using a calculator, \( \sin(5.4^{\circ}) \approx 0.094 \).
05

Rearrange Equation

Rearrange the equation \( d\sin(5.4^{\circ}) = 1.5\lambda \) to solve for the ratio: \( \frac{d}{\lambda} = \frac{1.5}{\sin(5.4^{\circ})} \).
06

Calculate the Ratio

Using the value of \( \sin(5.4^{\circ}) \approx 0.094 \), substitute back into the equation: \( \frac{d}{\lambda} = \frac{1.5}{0.094} \approx 15.96 \).

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Key Concepts

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

Double-slit interference
Young's double-slit experiment is a classic demonstration of the wave nature of light. When coherent light passes through two closely spaced slits, it creates an interference pattern on a screen placed behind the slits. This pattern consists of alternating bright and dark bands, known as fringes. The phenomenon occurs because the light waves emerging from the slits spread out and overlap. As they overlap, they interfere with each other.
Constructive interference results in bright fringes where the wave peaks meet, and destructive interference leads to dark fringes where a wave peak and trough meet.
In simple terms:
  • Bright fringes occur due to constructive interference (waves add up).
  • Dark fringes occur due to destructive interference (waves cancel out).
The position and spacing of these fringes depend on factors such as the wavelength of the light and the separation between the slits. Understanding this interference is essential to mastering wave phenomena and fundamentals in optics.
Dark fringe analysis
Dark fringes are crucial in analyzing and understanding the interference pattern created in a double-slit experiment. They are the points where destructive interference causes the light intensity to become zero. The position of these dark fringes is determined by the equation: \[ d\sin(\theta) = (m+\frac{1}{2})\lambda \]
Here, \(d\) is the slit separation, \(\theta\) is the angle of the fringe from the center, \(m\) is the order of the fringe (an integer value), and \(\lambda\) is the wavelength of the light. For dark fringes:
  • The term \((m+\frac{1}{2})\) ensures the wave path difference is half a wavelength, resulting in destructive interference.
  • The angle \(\theta\) is the specific direction in which these dark regions appear relative to the direction of the incoming wave.
By understanding where these dark fringes form, one can infer details about the light's wavelength and the geometric properties of the setup. This analysis is a foundational tool in the study of wave optics.
Wavelength and slit separation ratio
The ratio \( \frac{d}{\lambda} \), where \(d\) is the slit separation and \(\lambda\) is the wavelength of light, is a critical factor influencing the double-slit interference pattern. This ratio helps us understand how the geometry of the setup and the physical properties of light interact.
From the relationship \( d\sin(\theta) = (m+\frac{1}{2})\lambda \), rearranging gives \( \frac{d}{\lambda} = \frac{m+\frac{1}{2}}{\sin(\theta)} \).
This reveals that:
  • A larger \(d/\lambda\) ratio indicates that the slits are relatively far apart compared to the wavelength, leading to more closely spaced fringes.
  • A smaller \(d/\lambda\) suggests that the fringes are wider apart, as the slits are closer together or the wavelength is longer.
This ratio is vital in predicting and adjusting the fringe visibility and spacing, providing insight into practical applications of interference in technology and fundamental research. Using known values like the angle for the dark fringe can accurately calculate this ratio, as illustrated in the original exercise solution.

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Most popular questions from this chapter

The light shining on a diffraction grating has a wavelength of \(495 \mathrm{nm}\) (in vacuum). The grating produces a second-order bright iringe whose position is defined by an angle of \(9.34^{\circ} .\) How many lines per centimeter does the grating have?

The distance between adjacent slits of a certain diffraction grating is \(1.250 \times 10^{-5} \mathrm{m} .\) The grating is illuminated by monochromatic light with a wavelength of \(656.0 \mathrm{nm},\) and is then heated so that its temperature increases by \(100.0 \mathrm{C}^{\circ} .\) Determine the change in the angle of the seventh-order principal maximum that occurs as a result of the thermal expansion of the grating. The coefficient of linear expansion for the diffraction grating is \(1.30 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1} .\) Be sure to include the proper algebraic sign with your answer: \(+\) if the angle increases, \(-\) if the angle decreases.

The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen and the slit. Find the ratio \(\lambda / W\) of the wavelength \(\lambda\) of the light to the width \(W\) of the slit.

A circular drop of oil lies on a smooth, horizontal surface. The drop is thickest in the center and tapers to zero thickness at the edge. When illuminated from above by blue light \((\lambda=455 \mathrm{nm}), 56\) concentric bright rings are visible, including a bright fringe at the edge of the drop. In addition, there is a bright spot in the center of the drop. When the drop is illuminated from above by red light \((\lambda=637 \mathrm{nm}),\) a bright spot again appears at the center, along with a different number of bright rings. Ignoring the bright spot, how many bright rings appear in red light? Assume that the index of refraction of the oil is the same for both wavelengths. The ability to exhibit interference effects is a fundamental characteristic of any kind of wave. Our understanding of these effects depends on the principle of linear superposition, which we first encountered in Chapter 17\. Only by means of this principle can we understand the constructive and destructive interference of light waves that lie at the heart of every topic in this chapter. Problem 67 serves as a review of the essence of this principle. Problem 68 deals with thin-film interference and reviews the factors that must be considered in such cases.

The same diffraction grating is used with two different wave-lengths of light, \(\lambda_{A}\) and \(\lambda_{\mathrm{B}} .\) The fourth-order principal maximum of light A exactly overlaps the third-order principal maximum of light B. Find the ratio \(\lambda_{N} / \lambda_{\mathrm{B}}\)

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