Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 34: Problem 23

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?

Short Answer

Expert verified
Answer: The distance between the slits and the screen is approximately 1.0 m.

Step by step solution

01

Write the formula for the position of the maxima in an interference pattern

Write the formula as \(y_n = \frac{n\lambda L}{d}\).
02

Plug in the given values

We know the following values: - Wavelength \(\lambda = 540 \,\text{nm} = 540 \times 10^{-9}\, \text{m}\) (converting to meters) - Separation between the slits \(d = 0.100\, \text{mm} = 0.100 \times 10^{-3} \,\text{m}\) (converting to meters) - Distance between the first side maximum and the center of the pattern \(y_1 = 5.40\, \text{mm} = 5.40 \times 10^{-3}\, \text{m}\) (converting to meters) - Order of the maximum \(n = 1\) Now plug these values into the formula: \(5.40\times 10^{-3}=\frac{1\times 540\times 10^{-9} L}{0.100\times 10^{-3}}\)
03

Solve for L

To find the value of \(L\), we will first multiply both sides of the equation by \(0.100\times 10^{-3}\): \(5.40\times 10^{-3} \cdot 0.100 \times 10^{-3} = 540 \times 10^{-9} L\) Now, divide both sides by the remaining factor on the right side of the equation, which in this case is \(540 \times 10^{-9}\): \(L = \frac{5.40 \times 10^{-3} \cdot 0.100\times 10^{-3}}{540\times 10^{-9}}\) Upon calculating, we get: \(L \approx 1.0\,\text{m}\) Thus, the distance between the slits and the screen is approximately 1 meter.

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

In a double-slit arrangement the slits are \(1.00 \cdot 10^{-5} \mathrm{~m}\) apart. If light with wavelength \(500 .\) nm passes through the slits, what will be the distance between the \(m=1\) and \(m=3\) maxima on a screen \(1.00 \mathrm{~m}\) away?

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.

For a double-slit experiment, two 1.50 -mm wide slits are separated by a distance of \(1.00 \mathrm{~mm} .\) The slits are illuminated by a laser beam with wavelength \(633 \mathrm{nm} .\) If a screen is placed \(5.00 \mathrm{~m}\) away from the slits, determine the separation of the bright fringes on the screen.

What would happen to a double-slit interference pattern if a) the wavelength is increased? b) the separation distance between the slits is increased? c) the apparatus is placed in water?

A diffraction grating has \(4.00 \cdot 10^{3}\) lines \(/ \mathrm{cm}\) and has white light \((400 .-700 . \mathrm{nm})\) incident on it. What wavelength(s) will be visible at \(45.0^{\circ} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Magnetism

Read Explanation

Space Physics

Read Explanation

Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Electric Charge Field and Potential

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