/*! 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 24 Parallel rays of monochromatic l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 24

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 \(\mathrm{cm}\) from the slits. The centers of the slits are 0.640 \(\mathrm{mm}\) apart and the width of each slit is 0.434 \(\mathrm{mm}\) . If the intensity at the center of the central maximum is \(5.00 \times\) \(10^{-4} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity at a point on the screen that is 0.900 \(\mathrm{mm}\) from the center of the central maximum?

Short Answer

Expert verified
The intensity at 0.900 mm from the center is calculated using the interference intensity formula.

Step by step solution

01

Understand the Problem

We have a double-slit interference setup where we need to find the intensity at a point on the screen which is 0.900 mm away from the central maximum, given various parameters like slit separation, wavelength of light, distance to screen, and central maximum intensity.
02

Use the Double-Slit Interference Formula

In double-slit interference, the intensity at a point can be calculated using the formula:\[ I = I_0 \cos^2\left(\frac{\pi d x}{\lambda L}\right) \]where \(I_0\) is the intensity at the central maximum, \(d\) is the distance between slits, \(x\) is the distance of the point from the central maximum on the screen, \(\lambda\) is the wavelength of light, and \(L\) is the distance to the screen.
03

Substitute Known Values

Substitute the known values into the formula:- \(I_0 = 5.00 \times 10^{-4} \, \mathrm{W/m^2}\)- \(d = 0.640 \, \mathrm{mm} = 0.640 \times 10^{-3} \, \mathrm{m}\)- \(x = 0.900 \, \mathrm{mm} = 0.900 \times 10^{-3} \, \mathrm{m}\)- \(\lambda = 568 \, \mathrm{nm} = 568 \times 10^{-9} \, \mathrm{m}\)- \(L = 75.0 \, \mathrm{cm} = 0.750 \, \mathrm{m}\)Plugging these values into the equation:\[ I = (5.00 \times 10^{-4}) \cos^2\left(\frac{\pi (0.640 \times 10^{-3}) (0.900 \times 10^{-3})}{568 \times 10^{-9} \times 0.750}\right) \]
04

Calculate Angle

Calculate the angle in the cosine function:- First, simplify the expression inside the cosine:\[ \theta = \frac{\pi (0.640 \times 10^{-3}) (0.900 \times 10^{-3})}{568 \times 10^{-9} \times 0.750} \]- This evaluates to \(\theta\). Compute \(\theta\) using a calculator to find the angle in radians.
05

Compute the Cosine

Once \(\theta\) is computed, find \(\cos^2(\theta)\). Ensure the calculator is set to radians when computing the cosine value.
06

Calculate Intensity

Substitute \(\cos^2(\theta)\) back into the intensity equation to find:- \( I = (5.00 \times 10^{-4}) \cos^2(\theta) \) Evaluate this expression to find the final intensity at 0.900 mm from the central maximum.

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 91影视!

Key Concepts

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

Interference Pattern
An interference pattern is a fascinating phenomenon that occurs when waves overlap and combine. In the context of light, this often happens when monochromatic light passes through a double-slit. As each slit acts like a new source of waves, the light waves spread out and meet on a screen. At some points, the waves will add up (constructive interference) to make bright bands, while at others, they cancel out (destructive interference) leading to dark bands. This creates the characteristic pattern of alternating dark and bright bands, known as an interference pattern.

Understanding this pattern is key in experiments like double-slit setups because it helps illustrate the wave nature of light. By observing these patterns, scientists have gained deeper insights into the fundamental behaviors of waves.
Monochromatic Light
Monochromatic light refers to light that has a single wavelength or color. In practice, light sources such as lasers are often used to produce monochromatic light, as they emit light with just one wavelength. In the exercise, the wavelength of the monochromatic light is given as 568 nm, indicating the specific distance between one light wave crest and the next.

Using monochromatic light in double-slit interference is crucial because it ensures the clarity and precision of the interference pattern. With light of multiple wavelengths (like white light), different colors would create overlapping patterns, making it harder to analyze the results. Hence, monochromatic light helps in producing a well-defined and stable interference pattern.
Wavelength
Wavelength is a critical concept in understanding wave phenomena, including light waves. It is defined as the distance between consecutive peaks (or troughs) of a wave. In this problem, the wavelength of the light used is 568 nm, which is part of the visible spectrum and corresponds to a greenish-yellow color.

The role of wavelength in interference is significant as it influences the spacing of the interference pattern. The equation \[ I = I_0 \cos^2\left(\frac{\pi d x}{\lambda L}\right) \] shows how the wavelength \(\lambda\) is directly involved in calculating where the intensity maxima and minima occur on the screen. Shorter wavelengths result in patterns that have more closely spaced fringes, while longer wavelengths spread the fringes farther apart. This is why precise measurements of wavelength are crucial in many optical experiments.

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

In a large vacuum chamber, monochromatic laser light passes through a narrow slit in a thin aluminum plate and forms a diffraction pattern on a screen that is 0.620 \(\mathrm{m}\) from the slit. When the aluminum plate has a temperature of \(20.0^{\circ} \mathrm{C},\) the width of the central maximum in the diffraction pattern is 2.75 \(\mathrm{mm}\) . What is the change in the width of the central maximum when the temperature of the plate is raised to \(520.0^{\circ} \mathrm{C}\) ? Does the width of the central diffraction maximum increase or decrease when the temperature is increased?

If a diffraction grating produces its third-order bright band at an angle of \(78.4^{\circ}\) for light of wavelength \(681 \mathrm{nm},\) find (a) the number of slits per centimeter for the grating and (b) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

An interference pattern is produced by four parallel and equally spaced, narrow slits. By drawing appropriate phasor diagrams, show that there is an interference minimum when the phase difference \(\phi\) from adjacent slits is (a) \(\pi / 2 ;\) (b) \(\pi ;(\mathrm{c}) 3 \pi / 2\) . In each case, for which pairs of slits is there totally destructive interference?

Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0\(\mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{cm})\) of the eye to a height of about 50\(\mu \mathrm{m} .\) To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50 -\mum-tall object at 25 \(\mathrm{cm}\) from the eye with light of wavelength 550 \(\mathrm{nm}\) ? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25-\mathrm{cm}\) near point with light of wavelength 550 \(\mathrm{nm} ?(\mathrm{c})\) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \min =1^{\circ}\right),\) and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

An interference pattern is produced by light of wave- length 580 \(\mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 \(\mathrm{mm}\) . (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit, interference maxima? (b) Let the slits have width 0.320 \(\mathrm{mm}\) . In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Astrophysics

Read Explanation

Fluids

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Classical Mechanics

Read Explanation
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.