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Chapter 27: Problem 55

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure 27.6a. The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times\) \(10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm}\) ) of the monochromatic light?

Short Answer

Expert verified
The wavelength is 571.25 nm.

Step by step solution

01

Understand the Problem

In a Young's double-slit experiment, we have two light rays meeting at a point on a screen to form an eighth-order bright fringe. We are given the path difference as \(4.57 \times 10^{-6} \mathrm{m}\) and need to find the wavelength of the light.
02

Recall the Formula

In double-slit experiments, the condition for the bright fringes (constructive interference) is given by \( d \sin \theta = m\lambda \). Here, \( m \) is the order of the fringe, and \( \lambda \) is the wavelength. In this case, the order \( m \) is 8 and the path difference is \( m\lambda \).
03

Plug in the Given Values

Given the order of the bright fringe \( m = 8 \) and the path difference is \( 4.57 \times 10^{-6} \mathrm{m} \). According to the bright fringe formula, \( 4.57 \times 10^{-6} = 8\lambda \).
04

Solve for Wavelength

To find \( \lambda \), rearrange the equation to \( \lambda = \frac{4.57 \times 10^{-6}}{8} \). Calculate the wavelength: \[ \lambda = \frac{4.57 \times 10^{-6}}{8} = 5.7125 \times 10^{-7} \mathrm{m} \].
05

Convert the Wavelength to Nanometers

Convert the wavelength from meters to nanometers by multiplying by \(10^9\). \( 5.7125 \times 10^{-7} \mathrm{m} \times 10^9 = 571.25 \mathrm{nm} \).

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Key Concepts

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

Interference Pattern
In Young's double-slit experiment, understanding how the interference pattern forms is essential. It involves two light waves emerging from closely spaced slits and overlapping. This overlap creates patterns of bright and dark bands on a screen.
The bright spots, or fringes, occur due to constructive interference, where peaks of light waves align. The dark areas, on the contrary, are results of destructive interference, where peaks align with troughs.
These interference patterns are not just beautiful displays鈥攖hey demonstrate the wave nature of light and underscore principles like diffraction and superposition. The separation and distinctness of fringes can vary based on factors like the wavelength of the light used and the distance between the slits.
Wavelength Calculation
Calculating the wavelength in Young's double-slit experiment may seem complex at first, but it is straightforward with the correct formula. The key is to use the formula for constructive interference. In this context, it is given by:
  • \( d \sin \theta = m\lambda \)
  • \( m \) stands for the order of the bright fringe (e.g., 8th-order for this problem).
  • \( \lambda \) is what we solve for: the wavelength of the light.

We rearrange this equation considering that \( m\lambda \) equals the path difference between the two waves meeting on the screen. Given a path difference of \( 4.57 \times 10^{-6} \text{ m} \), we solve:\[ \lambda = \frac{4.57 \times 10^{-6}}{8} \]Plugging in these values, the wavelength calculates to \( 5.7125 \times 10^{-7} \text{ m} \), which converts to 571.25 nm when switched to nanometers.
Constructive Interference
Constructive interference is a fascinating phenomenon in Young's double-slit experiment. It occurs when the path difference between two overlapping waves on a screen equals a whole number of wavelengths (i.e., \( m\lambda \)).
This results in the bright fringes, as the crests of one wave align perfectly with the crests of another. It enhances the light's intensity at that point.
In formula terms, constructive interference appears when:
  • The equation \( d \sin \theta = m\lambda \) holds true.
  • Where \( m \) denotes the order of the fringe鈥攁 large integer indicating the bright fringe's position.
  • \( \lambda \) is the wavelength. It plays a key role as different wavelengths will shift the positions of these fringes.
Grasping this principle is vital, as constructive interference explains why certain light patterns emerge distinctly in optical experiments. It is the reason behind the aesthetically pleasing and scientifically valuable interference patterns seen in these setups.

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Most popular questions from this chapter

Crude Thickness Monitor. You and your team need to determine the thicknesses of two extremely thin sheets of plastic and do not have a measurement instrument capable of the job. You find a white light source with color filters (i.e., tinted glass windows that only let certain colors pass through) including red \((\lambda=660 \mathrm{nm}),\) green \((\lambda=550 \mathrm{nm}),\) and blue \((\lambda=\) \(470 \mathrm{nm}) .\) You locate two glass microscope slides, each of length \(10.0 \mathrm{cm}\) stack one on top of the other, and tape their ends together on one side. On the very edge of the side opposite the tape, you sandwich a piece of plastic between the slides to form a wedge of air between the plates. (a) When you shine blue light perpendicular to the slides, 55 bright fringes form along the full length of the slide. How thick is the plastic? (b) When you replace the first sheet with the second sheet of plastic, the fringes that appear are too closely spaced to count. You switch to red light (a longer wavelength), and count 124 fringes. How thick is the plastic? (c) How many fringes would you expect in each case, (a) and (b), if you instead used green light?

Light of wavelength \(410 \mathrm{nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(1.2 \times 10^{-5} \mathrm{m} .\) The distance between the grating and the viewing screen is \(0.15 \mathrm{m} .\) A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (Hint: The diffraction angles are small enough that the approximation \(\tan \theta \approx \sin \theta\) can be used.) (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right),\) what is the distance \(y ?\)

A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{m} .\) When light with a wavelength of \(\lambda_{1}=510 \mathrm{nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ), and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\)

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is \(675 \mathrm{nm} .\) Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{m}\) and (b) \(1.8 \times 10^{-6} \mathrm{m}\)

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 \(7.0 \mathrm{mm} .\) The taillights of this car are separated by a distance of \(1.2 \mathrm{m}\) 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?
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