/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Two slits spaced 0.450 \(\mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 \(\mathrm{nm} ?\)

Short Answer

Expert verified
The distance between the second and third dark lines is approximately 0.417 mm.

Step by step solution

01

Understand the Problem

We have two slits producing an interference pattern on a screen. We are asked to find the distance between the second and third dark fringes when coherent light of 500 nm wavelength is used. The key is to determine the position of these dark fringes using the formula for dark fringe positions in a double-slit experiment.
02

Identify the Formula

For dark fringes in a double-slit interference pattern, the condition is given by \( d \sin \theta = (m + 0.5) \lambda \), where \(d\) is the distance between the slits, \(\lambda\) is the wavelength, and \(m\) is the order of the dark fringe.
03

Calculate Positions of Dark Fringes

To find the positions, consider small angles where \( \sin \theta \approx \tan \theta \approx \frac{y}{L} \). Hence, \( y = (m + 0.5) \frac{\lambda L}{d} \). Substitute \( d = 0.450 \, \mathrm{mm} = 0.00045 \, \mathrm{m} \), \( \lambda = 500 \, \mathrm{nm} = 500 \times 10^{-9} \, \mathrm{m} \), and \( L = 0.75 \, \mathrm{m} \).
04

Calculate Position of Second Dark Line

Substitute \( m = 1 \) (second dark fringe), we get \( y_2 = (1 + 0.5) \frac{500 \times 10^{-9} \times 0.75}{0.00045} \). Calculate \( y_2 \).
05

Calculate Position of Third Dark Line

Substitute \( m = 2 \) (third dark fringe), we get \( y_3 = (2 + 0.5) \frac{500 \times 10^{-9} \times 0.75}{0.00045} \). Calculate \( y_3 \).
06

Find the Distance Between Dark Lines

The distance \( \Delta y \) between the second and third dark lines is \( y_3 - y_2 \). Substitute the calculated values of \( y_2 \) and \( y_3 \) to find \( \Delta y \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Dark Fringes in Double-Slit Interference
Dark fringes are key elements in understanding interference patterns created by light. In a double-slit experiment, light waves from each slit overlap, creating regions of both constructive and destructive interference. Dark fringes occur where destructive interference happens - spots where the waves from the two slits cancel each other out.

In simpler terms, a dark fringe is like a shadow on the bright background of light. These fringes occur at specific positions along the interference pattern on a screen.

The condition for dark fringes is given by the formula: \[ d \sin \theta = (m + 0.5) \lambda \]
  • \(d\) is the slit separation (0.450 mm in our problem).
  • \(\lambda\) is the wavelength of the light used (500 nm here).
  • \(m\) represents the order of the dark fringe (like second, third, etc.).
To find the locations of these fringes on a distant screen, we assume small angles where \(\sin \theta \approx \tan \theta \approx \frac{y}{L}\), letting us calculate the position \(y\) of dark fringes.
Wavelength and Its Role in Interference
Wavelength is a fundamental aspect of wave physics, representing the distance between successive peaks of a wave. It plays a crucial role in interference patterns, determining how waves interact to form bright and dark fringes.

The wavelength, \(\lambda\), is especially important in a double-slit experiment. It dictates the distance between interference fringes.

Longer wavelengths result in fringes that are farther apart, while shorter wavelengths create fringes that are closer together.
  • For the exercise, the wavelength of coherent light is given as 500 nm (nanometers).
  • This value tells us the light is visible and helps to find accurate positions of the dark lines.
Understanding wavelength helps in predicting and calculating the exact fringe pattern observed on the screen, enhancing insights into how light behaves when passing through slits.
Understanding the Interference Pattern
Interference patterns result from the interaction of overlapping waves, leading to regions of alternating light and dark stripes or bands on a screen. In the case of a double-slit experiment, the pattern stems from two sources of light waves interfering with each other.

Depending on where you look on the screen, you will either see a bright spot (where constructive interference occurs) or a dark fringe (where destructive interference occurs).
  • Constructive interference happens when light waves align to amplify the light, producing bright areas.
  • Destructive interference occurs when waves cancel each other out, leading to dark fringes.
Insight into this pattern aids in understanding fundamental wave properties, offering a window into the wave nature of light. The positions and spacing of these fringes are governed by the light’s wavelength, the distance between the slits, and the distance to the screen.
Coherent Light in Interference Experiments
Coherent light is essential in performing double-slit interference experiments effectively. It refers to light waves that are in phase and have a constant amplitude. Such light is characterized by having a fixed phase relationship, which means that the peaks and troughs of the waves match up over time.

In our exercise, the coherent light ensures the waves emerging from the slits maintain a consistent relationship. This consistency is key to forming clear interference patterns on the screen.
  • Lasers are a common source of coherent light due to their monochromatic and phase-aligned nature.
  • Other light sources can also be made coherent using special filters or apparatus.
With coherent light, you can accurately predict and measure the positions of dark and bright fringes, such as determining the distance between specific dark fringes like the second and third ones in this case.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) this film is just thick enough for light with wavelength 582.4 \(\mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength 588.5 \(\mathrm{nm}\) . What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04\(\mu \mathrm{m}\) apart, and in line with an observer, so that one source is 2.04\(\mu \mathrm{m}\) farther from the observer than the other. (a) For what visible wavelengths \((380\) to 750 \(\mathrm{nm})\) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04\(\mu \mathrm{m}\) farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

Reflective Coatings and Herring. Herring and related fish have a brilliant silvery appearance that camouflages them while they are swimming in a sunlit ocean. The silveriness is due to platelets attached to the surfaces of these fish. Each platelet is made up of several alternating layers of crystalline guanine \((n=1.80)\) and of cytoplasm \((n=1.333,\) the same as water), with a guanine layer on the outside in contact with the surrounding water (Fig. \(\mathrm{P} 35.56\) ). In one typical platelet, the guanine layers are 74 nm thick and the cytoplasm layers are 100 \(\mathrm{nm}\) thick. (a) For light striking the platelet surface at normal incidence, for which vacuum wavelengths of visible light will all of the reflections \(R_{1}\) , \(R_{2}, R_{3}, R_{4},\) and \(R_{5},\) shown in Fig. P35.56, be approximately in phase? If white light is shone on this platelet, what color will be most strongly reflected (see Fig. 32.4\() ?\) The surface of a herring has very many platelets side by side with layers of different thickness, so that all visible wavelengths are reflected. (b) Explain why such a "stack" of layers is more reflective than a single layer of guanine with cytoplasm underneath. (A stack of five guanine layers separated by cytoplasm layers reflects more than 80\(\%\) of incident light at the wavelength for which it is "tuned.") (c) The color that is most strongly reflected from a platelet depends on the angle at which it is viewed. Explain why this should be so. (You can see these changes in color by examining a herring from different angles. Most of the platelets on these fish are oriented in the same way, so that they are vertical when the fish is swimming.)

Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and 600 Hz. The speed of sound is 340 \(\mathrm{m} / \mathrm{s} .\) You find that where you are standing, you hear minimum- intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 \(\mathrm{cm}\) toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

Two speakers, 2.50 \(\mathrm{m}\) apart, are driven by the same audio oscillator so that each one produces a sound consisting of two distinct frequencies, 0.900 \(\mathrm{kHz}\) and 1.20 \(\mathrm{kHz}\) . The speed of sound in the room is 344 \(\mathrm{m} / \mathrm{s}\) . Find all the angles relative to the usual centerline in front of (and far from) the speakers at which both frequencies interfere constructively.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.