/*! 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 34 Light of wavelength 589 nm incid... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 34

Light of wavelength 589 nm incident on a pair of slits produces an interference pattern on a distant screen in which the separation between adjacent bright fringes at the center of the pattern is \(0.530 \mathrm{cm} .\) A second light source, when incident on the same pair of slits, produces an interference pattern on the same screen with a separation of $0.640 \mathrm{cm}$ between adjacent bright fringes at the center of the pattern. What is the wavelength of the second source? [Hint: Is the small-angle approximation justified?]

Short Answer

Expert verified
In a double-slit interference experiment, the separation between adjacent bright fringes is determined by the difference in the position between consecutive orders and is given by the formula \(\Delta x = \dfrac{\lambda D}{d}\), where \(\lambda\) is the wavelength of the light, \(D\) is the distance from the screen to the slits, and \(d\) is the distance between the slits. Using the given separations between adjacent bright fringes for the first and second light sources, we can find the ratio of their wavelengths and solve for the unknown wavelength of the second light source. The wavelength of the second light source is found to be 711 nm.

Step by step solution

01

Understand the double-slit interference formula

The general formula for the position of bright fringes in a double-slit interference experiment is given by: \(x = \dfrac{m\lambda D}{d}\) where \(x\) is the position of the bright fringe on the screen, \(m\) is the order of the fringe (an integer), \(\lambda\) is the wavelength of the light, \(D\) is the distance from the screen to the slits, and \(d\) is the distance between the slits.
02

Find the relationship between the separations of adjacent bright fringes and the formula

The separation between adjacent bright fringes is determined by the difference in the position between consecutive orders, i.e., \((m+1)-m\). Using the formula from Step 1, the separation is given by: \(\Delta x = x_{m+1} - x_m = \dfrac{(m+1)\lambda D}{d} - \dfrac{m\lambda D}{d} = \dfrac{\lambda D}{d}\)
03

Use the given data to find the ratio of the wavelengths

We are given the separation between adjacent bright fringes for the first and second light sources. We can use the relationship we found in Step 2 to find the ratio of their wavelengths: \(\dfrac{\Delta x_1}{\Delta x_2} = \dfrac{\lambda_1}{\lambda_2}\) Plugging in the given values: \(\dfrac{0.530}{0.640} = \dfrac{589}{\lambda_2}\)
04

Solve for the unknown wavelength

From the equation in Step 3, we can solve for the wavelength of the second light source (\(\lambda_2\)): \(\lambda_2 = \dfrac{589 \times 0.640}{0.530} = 711 \mathrm{nm}\) The wavelength of the second light source is 711 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

If diffraction were the only limitation, what would be the maximum distance at which the headlights of a car could be resolved (seen as two separate sources) by the naked human eye? The diameter of the pupil of the eye is about \(7 \mathrm{mm}\) when dark-adapted. Make reasonable estimates for the distance between the headlights and for the wavelength.
Find the height \(h\) of the pits on a CD (Fig. \(25.6 \mathrm{a}\) ). When the laser beam reflects partly from a pit and partly from land (the flat aluminum surface) on either side of the "pit," the two reflected beams interfere destructively; \(h\) is chosen to be the smallest possible height that causes destructive interference. The wavelength of the laser is \(780 \mathrm{nm}\) and the index of refraction of the poly carbonate plastic is \(n=1.55.\)
A spectrometer is used to analyze a light source. The screen-to-grating distance is \(50.0 \mathrm{cm}\) and the grating has 5000.0 slits/cm. Spectral lines are observed at the following angles: $12.98^{\circ}, 19.0^{\circ}, 26.7^{\circ}, 40.6^{\circ}, 42.4^{\circ}\( \)63.9^{\circ},\( and \)77.6^{\circ} .$ (a) How many different wavelengths are present in the spectrum of this light source? Find each of the wavelengths. (b) If a different grating with 2000.0 slits/cm were used, how many spectral lines would be seen on the screen on one side of the central maximum? Explain.
A Michelson interferometer is set up using white light. The arms are adjusted so that a bright white spot appears on the screen (constructive interference for all wavelengths). A slab of glass \((n=1.46)\) is inserted into one of the arms. To return to the white spot, the mirror in the other arm is moved $6.73 \mathrm{cm} .$ (a) Is the mirror moved in or out? Explain. (b) What is the thickness of the slab of glass?
Coherent light from a laser is split into two beams with intensities \(I_{0}\) and \(4 I_{0},\) respectively. What is the intensity of the light when the beams are recombined? If there is more than one possibility, give the range of possibilities. (tutorial: two waves).
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Measurements

Read Explanation

Electricity

Read Explanation

Wave Optics

Read Explanation

Further Mechanics and Thermal 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.