Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 34: Problem 40

Plane light waves are incident on a single slit of width \(2.00 \mathrm{~cm} .\) The second dark fringe is observed at \(43.0^{\circ}\) from the central axis. What is the wavelength of the light?

Short Answer

Expert verified
Answer: The wavelength of the light is approximately \(7.18 \times 10^{-7} \mathrm{~m}\).

Step by step solution

01

Determine the angle of the second dark fringe

First, we need to convert the given angle, \(43.0^{\circ}\), to radians, since we will be using the radians in subsequent calculations. To convert degrees to radians, we use the formula: $$ 1^{\circ} = \frac{\pi}{180} \: \text{radians} $$ Let \(\theta\) be the angle of the second dark fringe in radians. Then we have: $$ \theta = 43.0^{\circ} \times \frac{\pi}{180} = \frac{43\pi}{180} \: \text{radians} $$
02

Write the formula for the angular position of dark fringes

The formula for the angular position of the dark fringes in a single-slit diffraction pattern is given by: $$ \sin{\theta} = (m + \frac{1}{2})\frac{\lambda}{w} $$ Where: \(\sin \theta\) is the sine of the angle of the dark fringe \(m\) is the order of the dark fringe (0, 1, 2, 3,...) \(\lambda\) is the wavelength of the light \(w\) is the width of the single slit We know that this problem involves the second dark fringe, which means \(m = 1\). We also know the width of the single slit, \(w = 2.00 \mathrm{~cm}\). Now, we will solve for \(\lambda\).
03

Solve for the wavelength of the light

Plug the given values and the angle we found in radians into the formula of the angular position of dark fringes: $$ \sin{\frac{43\pi}{180}} = \left(1 + \frac{1}{2}\right)\frac{\lambda}{2.00 \mathrm{~cm}} $$ To solve for \(\lambda\), first, multiply both sides by \(2.00 \mathrm{~cm}\): $$ (2.00 \mathrm{~cm})\sin{\frac{43\pi}{180}} = (1 + \frac{1}{2})\lambda $$ Now, we divide both sides by \((1 + \frac{1}{2})\): $$ \lambda = \frac{(2.00 \mathrm{~cm})\sin(\frac{43\pi}{180})}{1+\frac{1}{2}} $$ Finally, plug in the known values and compute for \(\lambda\): $$ \lambda = \frac{(2.00 \mathrm{~cm})\sin(\frac{43\pi}{180})}{1.5} \approx 7.18 \times 10^{-7} \mathrm{~m} $$
04

State the final answer

The wavelength of the light causing the second dark fringe at \(43.0^{\circ}\) in a single-slit diffraction pattern with a slit width of \(2.00 \mathrm{~cm}\) is approximately \(7.18 \times 10^{-7} \mathrm{~m}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A glass with a refractive index of 1.50 is inserted into one arm of a Michelson interferometer that uses a 600.-nm light source. This causes the fringe pattern to shift by exactly 1000 fringes. How thick is the glass?

The Michelson interferometer is used in a class of commercially available optical instruments called wavelength meters. In a wavelength meter, the interferometer is illuminated simultaneously with the parallel beam of a reference laser of known wavelength and that of an unknown laser. The movable mirror of the interferometer is then displaced by a distance \(\Delta d,\) and the number of fringes produced by each laser and passing by a reference point (a photo detector) is counted. In a given wavelength meter, a red He-Ne laser \(\left(\lambda_{\mathrm{Red}}=632.8 \mathrm{nm}\right)\) is used as a reference laser. When the movable mirror of the interferometer is displaced by a distance \(\Delta d\), a number \(\Delta N_{\text {Red }}=6.000 \cdot 10^{4}\) red fringes and \(\Delta N_{\text {unknown }}=7.780 \cdot 10^{4}\) fringes pass by the reference photodiode. a) Calculate the wavelength of the unknown laser. b) Calculate the displacement, \(\Delta d\), of the movable mirror.

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?

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

A \(5.000-\mathrm{cm}\) -wide diffraction grating with 200 grooves is used to resolve two closely spaced lines (a doublet) in a spectrum. The doublet consists of two wavelengths, \(\lambda_{\mathrm{a}}=\) \(629.8 \mathrm{nm}\) and \(\lambda_{\mathrm{b}}=630.2 \mathrm{nm} .\) The light illuminates the entire grating at normal incidence. Calculate to four significant digits the angles \(\theta_{1 \mathrm{a}}\) and \(\theta_{1 \mathrm{~b}}\) with respect to the normal at which the first-order diffracted beams for the two wavelengths, \(\lambda_{\mathrm{a}}\) and \(\lambda_{\mathrm{b}}\), respectively, will be reflected from the grating. Note that this is not \(0^{\circ}\) What order of diffraction is required to resolve these two lines using this grating?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Fluids

Read Explanation

Electric Charge Field and Potential

Read Explanation

Geometrical and Physical Optics

Read Explanation

Space Physics

Read Explanation

Fundamentals of Physics

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.

Sign-up for free