/*! 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 60 After a laser beam passes throug... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) \(d/\lambda \approx 1.535\); (b) \(30.7^{\circ}\).

Step by step solution

01

Understanding the Problem

We have a two-slit interference problem where the dark fringes occur when the path difference between light from the two slits is a half-integer multiple of the wavelength, indicated as \(d \sin \theta = (m + \frac{1}{2})\lambda\). The first completely dark fringes occur at \(\pm 19.0^{\circ}\), which means \(m = 0\). We need to determine the ratio \(d/\lambda\).
02

Finding the Ratio of Slit Separation to Wavelength (Part a)

For dark fringes, the condition is \(d \sin \theta = (m + \frac{1}{2})\lambda\), where \(m = 0\) for the first minimum. Substituting \(\theta = 19.0^{\circ}\) and \(m = 0\), we get \(d \sin 19.0^{\circ} = \frac{1}{2}\lambda\). Solving for \(d/\lambda\), we have \(d/\lambda = \frac{1}{2\sin 19.0^{\circ}}\). Calculate \(\sin 19.0^{\circ}\) and find the ratio.
03

Calculation of \(d/\lambda\)

Calculate \(\sin 19.0^{\circ} \approx 0.3256\) (using a calculator). Then, \(d/\lambda = \frac{1}{2(0.3256)} \approx 1.535\).
04

Understanding Intensity Variation (Part b)

For the intensity to be \(\frac{1}{10}\) of the maximum, we need to use the intensity formula \(I = I_{max} \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right)\) and solve for \(\theta\) such that \(I = \frac{1}{10}I_{max}\). This gives \(\cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{10}\).
05

Solving for the Angle \(\theta\)

Solving the equation \(\cos^2 x = \frac{1}{10}\), we find \(x = \frac{\pi}{3}\approx 1.047\) radians or \(-\frac{\pi}{3}\). Therefore, \(\frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{3}\), so \(d \sin \theta/\lambda = 1/3\). Substitute \(d/\lambda = 1.535\) to solve for \(\sin \theta\), giving \(\sin \theta = \frac{1.535}{3} \approx 0.5117\). Find \(\theta\) such that \(\sin \theta \approx 0.5117\) results in \(\theta \approx 30.7^{\circ}\).
06

Conclusion

The ratio of distance between the slits to the wavelength is approximately 1.535, and the smallest angle where the light intensity is \(\frac{1}{10}\) of the maximum is approximately \(30.7^{\circ}\).

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 two-slit interference, dark fringes are areas on the screen where the light waves cancel each other out, resulting in no light or a "dark" spot. This occurs when the path difference between the light coming from the two slits equals a half-integer multiple of the wavelength, specifically \(d \sin \theta = (m + \frac{1}{2})\lambda\). In this formula, \(d\) represents the slit separation, \(\theta\) is the angle to the fringe, and \(\lambda\) denotes the wavelength of the light. When the laser light in our scenario passed through the slits, the first dark fringes were observed at \(\pm 19.0^{\circ}\). To comprehend this, keep in mind that dark fringes are analogous to quiet places in an echoing room, where sound waves block each other out instead of constructing sound.
Intensity Variation
The intensity variation in a two-slit interference pattern is determined by how the light's brightness changes from one point to another on the screen. The intensity is at its maximum where the light waves from the two slits constructively interfere and is significantly reduced, or even becomes zero, where they destructively interfere. The formula for light intensity in this context is \(I = I_{max} \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right)\). This equation indicates how the intensity \(I\) varies across angles \(\theta\) and shows that the maximum intensity \(I_{max}\) is modified by the cosine squared term. If we want to find out at which angle the intensity drops to \(\frac{1}{10}\) of its maximum, we solve \(\cos^2 x = \frac{1}{10}\), which helps us find where this lessened intensity occurs.
Wavelength Calculation
Wavelength calculation is crucial in experiments like the two-slit interference. It helps us understand how the light's color and type affect the interference pattern observed. For our experiment, knowing the ratio \(d/\lambda\) aids in deducing the wavelength of the light used, given the slit separation and fringe positions. By using equations such as \(d \sin \theta = (m + \frac{1}{2})\lambda\) for dark fringes, and working backwards, we can calculate or estimate the wavelength if \(d\) and the angle \(\theta\) are known. Accurately calculating the wavelength can reveal properties about the light source.
Slit Separation Ratio
The slit separation ratio, denoted as \(d/\lambda\), is a key parameter that affects the interference pattern in a two-slit experiment. It tells us how the physical spacing between the slits compares to the wavelength of the light passing through them. A higher ratio results in fringes that are closer together, while a lower ratio spreads them farther apart. In our given problem, we calculated \(d/\lambda\) to be approximately 1.535, meaning the slits are about 1.535 times the wavelength apart. This ratio provides insight into how finely or broadly the interference pattern is spread on the screen, which is essential for designing experiments and understanding wave propagation.

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

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction \(1.52 .\) (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ},\) what is the intensity? (b) What is the path difference for 480 -nm light from the two slits at a point where the phase angle is \(60.0^{\circ} ?\)

Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) (red) and 470 nm (blue), passes through two narrow slits separated by \(0.300 \mathrm{mm},\) and the interference pattern is observed on a screen 5.00 \(\mathrm{m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

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?

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.