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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 Dark Fringe Condition

The position of dark fringes in a double-slit experiment is given by the equation \( d \sin \theta = (m + 0.5) \lambda \), where \( m \) is an integer. The second dark fringe corresponds to \( m = 1 \).
02

Setting Up the Equation for Second Dark Fringe

For the second dark fringe, we use \( m = 1 \). Thus, the equation becomes \( d \sin 5.4^{\circ} = (1 + 0.5) \lambda = 1.5 \lambda \).
03

Solving for \( \frac{d}{\lambda} \)

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

Calculating the Ratio \( \frac{d}{\lambda} \)

Using \( \sin 5.4^{\circ} \approx 0.094 \), substitute to find \( \frac{d}{\lambda} \): \( \frac{d}{\lambda} = \frac{1.5}{0.094} \approx 15.96 \).

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Key Concepts

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

Dark Fringe Condition
In Young's double-slit experiment, understanding the position of dark fringes is crucial for interpreting the results. A dark fringe occurs due to destructive interference of light. This happens when the light waves from two slits are out of phase, which causes them to cancel each other out.
Destructive interference forms dark bands on a screen known as dark fringes. The condition for these dark fringes to occur is mathematically described by the formula:
  • \( d \sin \theta = (m + 0.5) \lambda \)
Here:
  • \(d\) is the distance between the slits.
  • \(\theta\) is the angle of diffraction.
  • \(m\) is an integer representing the order of the fringe.
  • \(\lambda\) is the wavelength of the light.
For the second dark fringe, \(m = 1\), making the formula \(d \sin \theta = 1.5 \lambda\). This formula helps to find which angles correspond to dark areas on the pattern, indicating where light cancels out.
Fringe Pattern
The fringe pattern observed in Young's double-slit experiment provides visual evidence of wave interference. The pattern consists of alternating bright and dark lines or bands. Bright fringes are formed by constructive interference where light waves from two slits add up, amplifying the light's intensity.
Dark fringes, on the other hand, are areas where the waves destructively interfere, cancelling each other out. The separation and number of these fringes depend on factors such as the wavelength of light and the distance between the slits.
The fringe pattern speaks to the wave-like nature of light and proves wave interference, being a repetitive yet predictable outcome of the experiment. Each fringe pattern is unique to the setup of the experiment, majorly determined by \(d\) and \(\lambda\).
Light Wavelength
The wavelength of light plays a vital role in Young's double-slit experiment as it determines the spacing of the fringes in the interference pattern. A shorter wavelength results in more closely spaced fringes, while a longer wavelength spreads them further apart.
Wavelength \(\lambda\) is essentially the distance between successive peaks of a wave and is a fundamental property of light, dictating its color in the visible spectrum. For example, blue light has a shorter wavelength compared to red light.
Knowing the wavelength is crucial in settings involving precise measurements and in thoroughly understanding the interference patterns formed in the experiment. It directly affects the formula used to calculate the positions of both bright and dark fringes in the pattern.
Angle of Diffraction
The angle of diffraction \(\theta\) describes the direction in which light spreads after passing through the double slits. This angle varies for different orders of bright and dark fringes in the interference pattern. The angle at which the second dark fringe appears provides key information about the interference and is vital for using formulas to calculate results like the slit separation and wavelength relations.
For a given fringe, this angle can be calculated using:
  • \(\sin \theta = \frac{m + 0.5}{\frac{d}{\lambda}}\)
As demonstrated in the provided solution, knowing the angle allows us to decipher the experiment's parameters, such as finding the ratio \(\frac{d}{\lambda}\). By understanding the angle of diffraction, scientists can predict and analyze the behavior of waves and apply this knowledge in various fields, from optical instruments to communication technologies.

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Most popular questions from this chapter

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 \(\mathrm{nm}\) . What is the vacuum wavelength of the light in \(\mathrm{nm}\) ?

A sheet that is made of plastic \((n=1.60)\) covers one slit of a double slit (see the drawing). When the double slit is illuminated by monochromatic light \(\left(\lambda_{\text { vacuum }}=586 \mathrm{nm}\right),\) the center of the screen appears dark rather than bright. What is the minimum thickness of the plastic?

In a single-slit diffraction pattern on a flat screen, the central bright fringe is 1.2 \(\mathrm{cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{m} .\) When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of 1.9 \(\mathrm{cm} .\) What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta\)

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about 7.0 mm. The taillights of this car are separated by a distance of 1.2 m and emit red light (wavelength \(=660 \mathrm{nm}\) in vacuum. How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

A uniform layer of water \((n=1.33)\) lies on a glass plate \((n=1.52) .\) Light shines perpendicularly on the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the light is 432 \(\mathrm{nm}\) in vacuum and also when it is 648 \(\mathrm{nm}\) in vacuum. \((\mathrm{a})\) Obtain the minimum thickness of the film. (b) Assuming that the film has the minimum thickness and that the visible spectrum extends from 380 to \(750 \mathrm{nm},\) determine the visible wavelength(s) in vacuum for which the film appears completely dark.

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