Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Two very narrow slits are spaced 1.80 \(\mu\)m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda\) = 550 nm? (Hint: The angle \(\theta\) in Eq. (35.5) is \(not\) small.)

Short Answer

Expert verified
The distance between the first and second dark lines is approximately 3.1 mm.

Step by step solution

01

Understand the Problem

We need to calculate the distance between the first and second dark lines in the interference pattern from two slits using the given wavelength \(\lambda = 550\ \text{nm}\), slit separation \(d = 1.80\ \mu\text{m}\), and distance to the screen \(L = 35\ \text{cm}\). The dark lines appear where destructive interference occurs, given by the path difference \(d\sin\theta = \left(m + \frac{1}{2}\right)\lambda\), where \(m\) is an integer for dark lines.
02

Use the Condition for Dark Lines

For the first dark line, \(m = 0\) so the condition becomes:\[d\sin\theta_1 = \left(0 + \frac{1}{2}\right)\lambda = \frac{1}{2}\lambda\]For the second dark line, \(m = 1\) so:\[d\sin\theta_2 = \left(1 + \frac{1}{2}\right)\lambda = \frac{3}{2}\lambda\]
03

Calculate the Angles \(\theta_1\) and \(\theta_2\)

Solve for \(\theta_1\) and \(\theta_2\) using the equations:\[\sin\theta_1 = \frac{\frac{1}{2}\lambda}{d} = \frac{275}{1800}\approx 0.1528\]\[\sin\theta_2 = \frac{\frac{3}{2}\lambda}{d} = \frac{825}{1800}\approx 0.4583\]Find \(\theta_1 = \arcsin(0.1528)\) and \(\theta_2 = \arcsin(0.4583)\).
04

Calculate Position of Dark Lines on the Screen

Use \(y = L\tan\theta\) to find the position \(y_1\) for the first dark line and \(y_2\) for the second:\[y_1 = 35\times\tan(\theta_1)\]\[y_2 = 35\times\tan(\theta_2)\]Compute \(y_1\) and \(y_2\).
05

Find the Distance Between the Dark Lines

The distance \(\Delta y\) between the first and second dark lines is:\[\Delta y = y_2 - y_1\]Calculate the difference between the two positions obtained in Step 4 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 Vaia!

Key Concepts

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

Destructive interference
When we talk about interference patterns, "destructive interference" is a crucial concept. This happens when two or more waves overlap in such a way that they cancel each other out. The high points of one wave meet the low points of another, resulting in reduced or zero amplitude. This effect is very easy to observe in light waves, with the result being areas that appear dark on a screen.

In the context of our double-slit experiment, destructive interference results in dark lines on the screen where the waves from the two slits perfectly cancel each other out. This occurs when the path difference between the two waves is a half-multiple of the wavelength, specifically \[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]where:
  • \(d\) is the distance between the slits,
  • \(\lambda\) is the wavelength of the light,
  • \(\theta\) is the angle of the dark fringe from the horizontal,
  • \(m\) is an integer (0, 1, 2, ...).
This formula helps us find where the darkness happens, indicating complete wave cancellation on our pattern.
Double-slit experiment
The double-slit experiment, first conducted by Thomas Young, is a famous demonstration of the wave properties of light. In this experiment, coherent light — light of a single wavelength — shines on a pair of closely-spaced slits. As the light passes through these slits, it spreads out or diffracts.

The resulting light waves overlap and interfere with each other, producing distinct patterns on a screen behind the slits. We see a series of bright and dark lines. Bright lines occur from constructive interference, while dark lines, where no light appears, result from destructive interference.

This experiment is not only foundational in understanding wave properties of light but also delves into the principles of superposition, where the resulting wave pattern is a sum of individual wave patterns. As in our problem, by knowing slit spacing and wavelength, interference patterns teach us about wave interactions on a detailed level.
Wavelength calculation
Wavelength calculation is an integral part of solving interference problems. Wavelength, denoted as \(\lambda\), measures the distance between consecutive peaks of a wave. It's a key parameter that determines where the interference pattern's bright and dark spots will appear.

In calculating where dark lines form in interference patterns (as in our example), we use the formula specific to destructive interference:\[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]Here, \(m\) is the order of the dark fringe, and knowing the position \(\theta\) helps us locate these dark spots precisely on the screen.

To find where these lines appear on the screen, converting angles to linear position using \( y = L\tan\theta \), where \(L\) is the distance from the slits to the screen, is crucial. All these calculations revolve around understanding wavelength, showing its central role in predicting wave interference patterns effectively.

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

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.

A plate of glass 9.00 cm long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 mm thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 nm. How many interference fringes are observed per centimeter in the reflected light?

Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 \(\mu\)m apart, and in line with an observer, so that one source is 2.04 \(\mu\)m farther from the observer than the other. (a) For what visible wavelengths (380 to 750 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 \(\mu\)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?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

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.

Sign-up for free