/*! 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 62 Coherent monochromatic light pas... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 62

Coherent monochromatic light passes through parallel slits and then onto a screen that is at a distance \(L=2.40 \mathrm{~m}\) from the slits. The narrow slits are a distance \(d=2.00 \cdot 10^{-5} \mathrm{~m}\) apart. If the minimum spacing between bright spots is \(y=6.00 \mathrm{~cm},\) find the wavelength of the light.

Short Answer

Expert verified
Answer: The wavelength of the light is 10 nm.

Step by step solution

01

Finding the angle 胃 using the minimum spacing between bright spots, y

We'll use the small angle approximation formula: $$ \theta \approx tan(\theta) = \frac{y}{L} $$ where \(L\) is the distance from the slits to the screen, and \(y\) is the minimum spacing between bright spots. Plugging the given values: $$ \theta = \frac{6.00 \times 10^{-2} \mathrm{~m}}{2.40 \mathrm{~m}} $$ Calculate the value of \(\theta\): $$ \theta = 0.025 $$
02

Finding the order m for the first minimum using the angle 胃

For the first minimum (dark spot) in the interference pattern, the path difference between two rays is half a wavelength. Therefore, the corresponding angle will be: $$ \sin(\theta') = \frac{\lambda}{2d} $$ Using the small angle approximation and knowing that \(\theta' = \theta\), $$ \theta' \approx \frac{\lambda}{2d} $$ We can now find the order of the bright spot \(m\) by multiplying both sides of the equation by \(2\): $$ m = 2\theta' = \frac{\lambda}{d} $$ By using the value of \(\theta\) calculated in Step 1, $$ m = 2 \times 0.025 = 0.05 $$
03

Finding the wavelength 位 using the constructive interference formula

Now, we can use the constructive interference formula to find the wavelength of the light: $$ \sin(\theta) = \frac{m\lambda}{d} $$ Plugging the calculated values for \(m\), \(d\), and \(\theta\): $$ 0.025 = \frac{0.05\lambda}{2.00 \times 10^{-5} \mathrm{~m}} $$ Solve for \(\lambda\): $$ \lambda = \frac{0.025 \times 2.00 \times 10^{-5} \mathrm{~m}}{0.05} $$ Calculate the value of \(\lambda\): $$ \lambda = 1.00 \times 10^{-5} \mathrm{~m} $$ So, the wavelength of the light is \(1.00 \times 10^{-5} \mathrm{~m}\) or \(10\,\mathrm{nm}\).

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!

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

Coherent, monochromatic light of wavelength \(450.0 \mathrm{nm}\) is emitted from two locations and detected at another location. The path difference between the two routes taken by the light is \(20.25 \mathrm{~cm}\). Will the two light waves interfere destructively or constructively at the detection point?

White light shines on a sheet of mica that has a uniform thickness of \(1.30 \mu \mathrm{m} .\) When the reflected light is viewed using a spectrometer, it is noted that light with wavelengths of \(433.3 \mathrm{nm}, 487.5 \mathrm{nm}, 557.1 \mathrm{nm}, 650.0 \mathrm{nm}\), and \(780.0 \mathrm{nm}\) is not present in the reflected light. What is the index of refraction of the mica?

A Newton's ring apparatus consists of a convex lens with a large radius of curvature \(R\) placed on a flat glass disc. (a) Show that the distance \(x\) from the center to the air, thickness \(d,\) and the radius of curvature \(R\) are given by \(x^{2}=2 R d\) (b) Show that the radius of nth constructive interference is given by \(x_{\mathrm{n}}=\left[\left(n+\frac{1}{2}\right) \lambda R\right]^{1 / 2} .\) (c) How many bright fringes may be seen if it is viewed by red light of wavelength 700\. \(\mathrm{nm}\) for \(R=10.0 \mathrm{~m},\) and the plane glass disc diameter is \(5.00 \mathrm{~cm} ?\)

Many astronomical observatories, and especially radio observatories, are coupling several telescopes together. What are the advantages of this?

A two-slit apparatus is covered with a red \((670 \mathrm{nm})\) filter. When white light is shone on the filter, on the screen beyond the two-slit apparatus, there are nine interference maxima within the 4.50 -cm-wide central diffraction maximum. When a blue \((450 \mathrm{nm})\) filter replaces the red, how many interference maxima will there be in the central diffraction maximum, and how wide will that diffraction maximum be?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Nuclear Physics

Read Explanation

Linear Momentum

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.