FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 23, Problem 107QAP
To determine

(a)

Three consecutive angles of dark ring from the central bright spot on the retina of the eye

Expert Solution
Check Mark

Answer to Problem 107QAP

1st ring occurs at θ1=3.05×104rad=0.0175o

2nd ring occurs at θ2=5.58×104rad=0.0319o

3rd ring occurs as θ1=8.10×104rad=0.0464o

Explanation of Solution

Given:

Diameter of pupil = dpupil=2mm=2×103m

Wavelength of light = λ=500nm=5×107m

Formula used:

Angle of resolution is defined as,

  θ=mλd(in radian)for first order, m = 1.22for second order, m = 2.233for the third order ,  m = 3.238

Calculation:

From above formula,

  θ=1.22λdfor first order,θ1=1.22( 5× 10 7 m 2× 10 3 )=3.05× 10 4rad_=3.05×104rad( 180 o π)= 0.175o_for second order, m = 2.233, then, θ2=2.233( 5× 10 7 m 2× 10 3 )=5.58× 10 4rad_=5.58×104rad( 180 o π)= 0.0319o_for third order, m = 3.238,θ3=3.238( 5× 10 7 m 2× 10 3 )=8.10× 10 4rad_=8.10×104rad( 180 o π)= 0.0464o_

Conclusion:

Thus, from above discussion, we find the angle of resolution for three consecutive dark rings as,

1st ring occurs at θ1=3.05×104rad=0.0175o

2nd ring occurs at θ2=5.58×104rad=0.0319o

3rd ring occurs as θ1=8.10×104rad=0.0464o

To determine

(b)

The radius of dark rings from the bright spot

Expert Solution
Check Mark

Answer to Problem 107QAP

The radius of consecutive three rings are found as,

For 1st ring, x1=7.6×103mm

For 2nd ring, x2=1.4×102mm

For 3rd ring, x3=2.0×102mm

Explanation of Solution

Given:

Angle of resolution for consecutive three dark rings is as,

1st ring occurs at θ1=3.05×104rad=0.0175o

2nd ring occurs at θ2=5.58×104rad=0.0319o

3rd ring occurs as θ1=8.10×104rad=0.0464o

Distance from retina to eye lens = D=25mm

Formula used:

  x=θDHere,θ is in radian.

Calculation:

  For first ring, x1=θ1D=3.05×104rad×25mm=7.6×103mmfor second rings,x2=θ2D=5.58×104rad×25mm=1.4×102mmfor third rings,x1=θ3D=8.10×104rad×25mm=2.0×102mm

Conclusion:

Thus, we have found the distance of dark rings from the central bright spot.

To determine

(c)

The explanation about the diffraction of light in our eye

Expert Solution
Check Mark

Answer to Problem 107QAP

We cannot observe dark and bright rings because angle of resolution is very small to distinguish the consecutive dark and bright rings for human.

Explanation of Solution

Introduction:

Angle of resolution is inversely proportional to the diameter of aperture of the opening and directly proportional to the wavelength of light. As the diameter of opening increases, angle of resolution will decrease.

From above discussion, we see that the angle of resolution is very small. So, locations of the dark and bright fringes are very close to each other. Human eye is not capable to distinguish these thin rings. They feel it as the continuous light.

Conclusion:

Thus, we do not observe such diffraction of light because of very closeness of the consecutive rings to each other.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Part A: kg (a) Water at 20 °C (p = 998.3 and v = 1 × 10-6 m²/s) flows through a galvanised m³ iron pipe (k = 0.15 mm) with a diameter of 25 mm, entering the room at point A and discharging at point C from the fully opened gate valve B at a volumetric flow rate of 0.003 m³/s. Determine the required pressure at A, considering all the losses that occur in the system described in Figure Q1. Loss coefficients for pipe fittings have been provided in Table 1. [25 marks] (b) Due to corrosion within the pipe, the average flow velocity at C is observed to be V2 m/s after 10 years of operation whilst the pressure at A remains the same as determined in (a). Determine the average annual rate of growth of k within the pipe. [15 marks] 4₁ Figure Q1. Pipe system Page 2 25 mm
For an independent study project, you design an experiment to measure the speed of light. You propose to bounce laser light off a mirror that is 53.5 km due east and have it detected by a light sensor that is 119 m due south of the laser. The first problem is to orient the mirror so that the laser light reflects off the mirror and into the light sensor. (a) Determine the angle that the normal to the mirror should make with respect to due west.(b) Since you can read your protractor only so accurately, the mirror is slightly misaligned and the actual angle between the normal to the mirror and due west exceeds the desired amount by 0.003°. Determine how far south you need to move the light sensor in order to detect the reflected laser light.
A mirror hangs 1.67 m above the floor on a vertical wall. A ray of sunlight, reflected off the mirror, forms a spot on the floor 1.41 m from the wall. Later in the day, the spot has moved to a point 2.50 m from the wall. (a) What is the change in the angle of elevation of the Sun, between the two observations?

Chapter 23 Solutions

FlipIt for College Physics (Algebra Version - Six Months Access)

Ch. 23 - Prob. 11QAPCh. 23 - Prob. 12QAPCh. 23 - Prob. 13QAPCh. 23 - Prob. 14QAPCh. 23 - Prob. 15QAPCh. 23 - Prob. 16QAPCh. 23 - Prob. 17QAPCh. 23 - Prob. 18QAPCh. 23 - Prob. 19QAPCh. 23 - Prob. 20QAPCh. 23 - Prob. 21QAPCh. 23 - Prob. 22QAPCh. 23 - Prob. 23QAPCh. 23 - Prob. 24QAPCh. 23 - Prob. 25QAPCh. 23 - Prob. 26QAPCh. 23 - Prob. 27QAPCh. 23 - Prob. 28QAPCh. 23 - Prob. 29QAPCh. 23 - Prob. 30QAPCh. 23 - Prob. 31QAPCh. 23 - Prob. 32QAPCh. 23 - Prob. 33QAPCh. 23 - Prob. 34QAPCh. 23 - Prob. 35QAPCh. 23 - Prob. 36QAPCh. 23 - Prob. 37QAPCh. 23 - Prob. 38QAPCh. 23 - Prob. 39QAPCh. 23 - Prob. 40QAPCh. 23 - Prob. 41QAPCh. 23 - Prob. 42QAPCh. 23 - Prob. 43QAPCh. 23 - Prob. 44QAPCh. 23 - Prob. 45QAPCh. 23 - Prob. 46QAPCh. 23 - Prob. 47QAPCh. 23 - Prob. 48QAPCh. 23 - Prob. 49QAPCh. 23 - Prob. 50QAPCh. 23 - Prob. 51QAPCh. 23 - Prob. 52QAPCh. 23 - Prob. 53QAPCh. 23 - Prob. 54QAPCh. 23 - Prob. 55QAPCh. 23 - Prob. 56QAPCh. 23 - Prob. 57QAPCh. 23 - Prob. 58QAPCh. 23 - Prob. 59QAPCh. 23 - Prob. 60QAPCh. 23 - Prob. 61QAPCh. 23 - Prob. 62QAPCh. 23 - Prob. 63QAPCh. 23 - Prob. 64QAPCh. 23 - Prob. 65QAPCh. 23 - Prob. 66QAPCh. 23 - Prob. 67QAPCh. 23 - Prob. 68QAPCh. 23 - Prob. 69QAPCh. 23 - Prob. 70QAPCh. 23 - Prob. 71QAPCh. 23 - Prob. 72QAPCh. 23 - Prob. 73QAPCh. 23 - Prob. 74QAPCh. 23 - Prob. 75QAPCh. 23 - Prob. 76QAPCh. 23 - Prob. 77QAPCh. 23 - Prob. 78QAPCh. 23 - Prob. 79QAPCh. 23 - Prob. 80QAPCh. 23 - Prob. 81QAPCh. 23 - Prob. 82QAPCh. 23 - Prob. 83QAPCh. 23 - Prob. 84QAPCh. 23 - Prob. 85QAPCh. 23 - Prob. 86QAPCh. 23 - Prob. 87QAPCh. 23 - Prob. 88QAPCh. 23 - Prob. 89QAPCh. 23 - Prob. 90QAPCh. 23 - Prob. 91QAPCh. 23 - Prob. 92QAPCh. 23 - Prob. 93QAPCh. 23 - Prob. 94QAPCh. 23 - Prob. 95QAPCh. 23 - Prob. 96QAPCh. 23 - Prob. 97QAPCh. 23 - Prob. 98QAPCh. 23 - Prob. 99QAPCh. 23 - Prob. 100QAPCh. 23 - Prob. 101QAPCh. 23 - Prob. 102QAPCh. 23 - Prob. 103QAPCh. 23 - Prob. 104QAPCh. 23 - Prob. 105QAPCh. 23 - Prob. 106QAPCh. 23 - Prob. 107QAPCh. 23 - Prob. 108QAPCh. 23 - Prob. 109QAPCh. 23 - Prob. 110QAPCh. 23 - Prob. 111QAPCh. 23 - Prob. 112QAPCh. 23 - Prob. 113QAPCh. 23 - Prob. 114QAP
Knowledge Booster
Background pattern image
Algebra
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Convex and Concave Lenses; Author: Manocha Academy;https://www.youtube.com/watch?v=CJ6aB5ULqa0;License: Standard YouTube License, CC-BY