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Chapter 27: Problem 7

Refer to Interactive Solution 27.7 at for help in solving this problem. In a Young's double-slit experiment the separation \(y\) between the second-order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of 425 nm. Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta .\) Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{nm}\)

Short Answer

Expert verified
The separation \(y\) is approximately \(0.0248 \, \text{m}\) for a wavelength of \(585 \) nm.

Step by step solution

01

Identify the given data for the first scenario

We are given that the separation between the second-order bright fringe and the central bright fringe is \( y = 0.0180 \, \text{m} \), and the wavelength of the light \( \lambda_1 = 425 \, \text{nm} \). These are the known values for our first setup.
02

Understand the formula for fringe separation

In Young's double-slit experiment, the separation between bright fringes depends on the formula: \[ y = \frac{m \cdot \lambda \cdot L}{d} \]where \( m \) is the order of the fringe (2 in this case, as we are dealing with the second-order fringe), \( \lambda \) is the wavelength of the light, \( L \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. The angle approximation \( \sin \theta \approx \tan \theta \) is used when these angles are small.
03

Solve for constant \( \frac{L}{d} \) using first conditions

We know the separation is \( 0.0180 \, \text{m} \) for \( \lambda_1 = 425 \, \text{nm} \). Arrange the formula as follows to solve for \( \frac{L}{d} \):\[ 0.0180 = \frac{2 \cdot 425 \times 10^{-9} \cdot L}{d} \]Rearranging gives:\[ \frac{L}{d} = \frac{0.0180}{2 \times 425 \times 10^{-9}} \]
04

Calculate \( \frac{L}{d} \)

Substitute in the numbers to get:\[ \frac{L}{d} = \frac{0.0180}{850 \times 10^{-9}} = \frac{0.0180}{850 \cdot 10^{-9}} \approx 21176.47 \, \text{m}^{-1} \] (rounded to two decimal places).
05

Solve the formula for the second scenario with \( \lambda_2 = 585 \, \text{nm} \)

Use the same formula for the new wavelength:\[ y = \frac{2 \cdot 585 \times 10^{-9} \cdot L}{d} \]Plug in the value of \( \frac{L}{d} \):\[ y = \frac{2 \cdot 585 \times 10^{-9} \cdot 21176.47}{1} \]
06

Calculate the new separation \( y \)

Substitute and calculate:\[ y = 2 \cdot 585 \times 10^{-9} \cdot 21176.47 \approx 0.0248 \, \text{m} \]So, the separation between the second-order bright fringe and the central fringe is approximately \( 0.0248 \, \text{m} \) for a wavelength of \( 585 \) nm.

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Key Concepts

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

Fringe Separation
In Young's double-slit experiment, fringe separation refers to the distance between successive bright or dark bands visible on a screen. More specifically, it is the distance from the central bright fringe, also known as the zeroth-order fringe, to another bright fringe of interest, which can be the first-order, second-order, or even higher order fringe. This separation occurs because when light waves pass through the two slits, they spread out and overlap to create an interference pattern.

The formula that describes fringe separation in a double-slit experiment is given by:
  • \[ y = \frac{m \cdot \lambda \cdot L}{d} \]
Where:
  • \( y \) is the fringe separation.
  • \( m \) is the order of the fringe (e.g., \( m = 1\) for first-order, \( m = 2\) for second-order).
  • \( \lambda \) is the wavelength of the light source.
  • \( L \) is the distance from the slits to the screen.
  • \( d \) is the distance between the two slits.
If the angles involved in measuring the fringes are small, the approximation \( \sin \theta \approx \tan \theta \) can be applied. This approximation simplifies the calculations without significantly affecting the accuracy, especially useful when screens are far from the slits.
Wavelength
Wavelength, represented by \( \lambda \), is a fundamental characteristic of waves, such as light. It refers to the physical length of one complete oscillation or cycle of the wave, and for our purposes, it's measured in nanometers (nm).

In the context of Young's double-slit experiment, the wavelength of light used dramatically affects the patterns and separations of the fringes formed on the screen. The longer the wavelength, the greater the fringe separation, as can be inferred from the fringe separation formula above.

To illustrate, consider light used in the experiment at different wavelengths:
  • A shorter wavelength, like 425 nm, results in closely spaced fringes.
  • A longer wavelength, for instance, 585 nm, causes the fringes to spread further apart.
This happens because a longer wavelength results in a larger difference in phase between light arriving at different points on the screen. As a result, the interference pattern shifts, increasing the separation between fringes. This property makes wavelength measurement critical when analyzing interference patterns and predicting the resultant spacing of fringes.
Second-order Fringe
In the realm of interference patterns, the order of a fringe refers to its position relative to the central bright fringe. The second-order fringe is the second bright band observed on either side of the central fringe. It arises when light paths differ by integer multiples of the wavelength, in this case, two wavelengths.

Understanding second-order fringes is vital because:
  • They provide insights into the entire wave interference process.
  • Their measurement can be used to determine the wavelength of the light, given other parameters.
The second-order fringe is formed when the path difference between light waves from the two slits equals twice the wavelength. Thus, it holds a path difference condition of \( 2\lambda \).

In experiments, second-order fringes allow researchers to fine-tune measurements and confirm the consistency of their results over greater distances on the experimental media, such as a screen. Recognizing and understanding these higher order fringes enhance the depth of study into wave phenomena and interference.

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Most popular questions from this chapter

In Young's experiment a mixture of orange light \((611 \mathrm{nm})\) and blue light \((471 \mathrm{nm})\) shines on the double slit. The centers of the first- order bright blue fringes lie at the outer edges of a screen that is located \(0.500 \mathrm{~m}\) away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in which direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen? It may be assumed that \(\theta\) is small, so that \(\sin \theta \approx \tan \theta\).

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 \(\sin \theta \approx \tan \theta\).

Light shines on a diffraction grating, and a diffraction pattern is produced on a viewing screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) From trigonometry, how is the distance \(y\) from the central bright fringe to the second-order bright fringe related to the diffraction angle \(\theta\) and the distance \(L\) between the grating and the screen? (b) From physics, how is \(\theta\) related to the order \(m\) of the bright fringe, the wavelength \(\lambda\) of the light, and the separation \(d\) between the slits? (c) In this problem, the angle \(\theta\) is small (less than a few degrees). When the angle is small, \(\tan \theta\) is approximately equal to \(\sin \theta\), or \(\tan \theta \approx \sin \theta\). Using this approximation, obtain an expression for \(y\) in terms of \(L, m, \lambda\), and \(d\). (d) If the entire apparatus in the drawing is submerged in water, would you expect the distance \(y\) to increase, decrease, or remain unchanged? Why? Light of wavelength \(480 \mathrm{~nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(5.0 \times 10^{-7} \mathrm{~m}\). The distance between the grating and the viewing screen is \(0.15 \mathrm{~m}\). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right)\), what is the distance \(y ?\) Be sure your answer is consistent with part (d) of the Concept Questions.

A hunter who is a bit of a braggart claims that, from a distance of \(1.6 \mathrm{~km}\), he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{~nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{~mm}\), the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of \(498 \mathrm{~nm}\).

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