COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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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.

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Chapter 23 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

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
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