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Chapter 35: Problem 14

Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by \(0.200 \mathrm{mm},\) and the interference pattern is observed on a screen 4.00 \(\mathrm{m}\) from the slits. (a) What is the width (in mm) of the central interference maximum? (b) What is the width of the first-order bright fringe?

Short Answer

Expert verified
(a) 8 mm; (b) 4 mm

Step by step solution

01

Understand the Young's Double Slit Experiment

The problem involves the interference pattern created by coherent light passing through two slits. The interference pattern consists of bright and dark fringes on a screen. We need to find the width of the central maximum and the first-order bright fringe.
02

Determine the Central Maximum Width

The central maximum is formed between the two first-order dark fringes. Each dark fringe is a path difference of half a wavelength off the path difference for a bright fringe. For small angles, the position of these fringes is given by: \( y_n = \frac{n \lambda L}{d} \).Here, for the first dark fringe, \( n=0.5 \) and \( n=-0.5 \), wavelength \( \lambda = 400 \, \mathrm{nm} = 400 \times 10^{-9} \, \mathrm{m} \), distance \( L = 4.00 \, \mathrm{m} \), and slit separation \( d = 0.200 \, \mathrm{mm} = 0.200 \times 10^{-3} \, \mathrm{m} \).
03

Calculate the Position for First-order Dark Fringes

Substituting in the formula, find the positions for \( n=0.5 \) and \( n=-0.5 \):\( y_{0.5} = \frac{0.5 \times 400 \times 10^{-9} \times 4 }{0.200 \times 10^{-3}} = 4 \times 10^{-3} \, \mathrm{m} \) and \( y_{-0.5} = -4 \times 10^{-3} \, \mathrm{m} \).
04

Find the Width of the Central Maximum

The width of the central maximum is the distance between these two positions, \( \text{Width} = y_{0.5} - y_{-0.5} = 8 \times 10^{-3} \, \mathrm{m} \) or \( 8 \, \mathrm{mm} \).
05

Determine the Width of the First-order Bright Fringe

The width of the first-order bright fringe is the distance between the first and second \(n=1\) and \(n=-1\) order dark fringes on one side:\( y_1 = \frac{1 \times 400 \times 10^{-9} \times 4 }{0.200 \times 10^{-3}} = 8 \times 10^{-3} \, \mathrm{m} \).Therefore, width of the first-order bright fringe on one side is \( y_1 - y_{0.5} = 8 \times 10^{-3} - 4 \times 10^{-3} = 4 \times 10^{-3} \, \mathrm{m} \) or \( 4 \, \mathrm{mm} \).
06

Verify and Summarize the Solution

We computed the width of the central maximum to be 8 mm, and for a first-order bright fringe, as measured from the middle of one fringe to the same position on the next, it can be understood by the symmetry as half of the central maximum width; hence, 4 mm. This contains the deviation between consecutive bright or dark peaks and midpoints.

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Key Concepts

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

Interference Pattern
The interference pattern is a mesmerizing phenomenon seen in Young's Double Slit Experiment. When coherent light passes through two closely spaced slits, it diffracts and overlaps on a screen, creating regions of alternating light and dark bands.

This pattern results from the principle of superposition, where the crests and troughs of light waves add together. When the crests line up with crests, we see bright bands called fringes. Conversely, when crests meet troughs, the waves cancel out, creating dark bands.
  • Bright Fringes: Constructive interference occurs where the paths of light differ by an integer multiple of wavelengths.
  • Dark Fringes: Destructive interference happens when the path difference is an odd half-wavelength multiple.
Understanding this pattern helps in distinguishing the different regions on the interference screen.
Coherent Light
In Young's Double Slit Experiment, the concept of coherent light is crucial to understanding how an interference pattern is formed. Coherent light refers to light waves that maintain a constant phase difference. This consistency is key to producing clear and stable interference patterns.

Coherent light is usually produced using lasers, as they provide light of a single wavelength and in a single phase across space and time. This uniformity ensures that when the light passes through the slits, it maintains its orderly wave pattern, necessary for consistent constructive and destructive interference.
Central Maximum
The central maximum is the brightest and widest part of the interference pattern in Young's Double Slit Experiment. This region appears directly opposite the light source, between the two slits.

To understand its size, consider how the light waves from the slits travel the same path length to this point on the screen, resulting in maximum constructive interference. This creates the broadest bright band compared to other fringes.
  • It spans from the first-order dark fringe on one side to the same on the opposite side.
  • In this exercise, its width is calculated as 8 mm using the formula: \[ y_n = \frac{n \lambda L}{d} \]at the specific conditions given.
Recognizing the central maximum helps in defining reference points for measuring other parts of the pattern.
First-Order Bright Fringe
First-order bright fringes are the next set of bright fringes found alongside the central maximum. They are the brightest fringes immediately adjacent to it and help define the structure of the interference pattern.

In terms of measurement, each first-order bright fringe occurs at a path difference of one full wavelength from the reference central bright fringe position.
  • To calculate their position, use the same interference formula as for the central maximum but adjust for the first order (n=1): \[ y_1 = \frac{1 \times 400 \times 10^{-9} \times 4 }{0.200 \times 10^{-3}} \]
  • The width of this fringe (about 4 mm in this exercise) is determined by finding the difference in position between its adjacent dark fringes.
This visual pattern distinguishes itself by showing how light behaves under diffraction and interference principles.

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Most popular questions from this chapter

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm 19.0^{\circ}\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wave-length of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(\frac{1}{10}\) the maximum intensity on the screen?

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 thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wavelength 525 nm that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 \(\mathrm{nm}\) per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 \(\mathrm{cm} .\) Point \(P\) is 4.86 \(\mathrm{m}\) from \(A\) and 5.24 \(\mathrm{m}\) from B. What is the phase difference at \(P\) between these two waves?
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