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
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Chapter 23, Problem 49QAP
To determine

(a)

Calculate the range of the index of refraction of the material for visible light.

Expert Solution
Check Mark

Answer to Problem 49QAP

Refractive index for visible light ranges from 1.50 to 1.57

Explanation of Solution

Given:

For a certain optical medium the speed of light varies from a low value of  1.9×108 m/s for violet light to a high value of  2.00×108 m/s for red light.

Concept Used:

The relationship between refractive index and the speed of light is given as follows:

  n=cv

Where

  n:Refractive indexc:Speed of light in vaccuum = 3.00×108 m/sv:Speed of light in medium

Calculation:

As per the given problem,

Speed of violet light, vviolet=1.9×108 m/s

Speed of red light, vred= 2.00×108 m/s

Refractive index corresponding to violet light nvioletis calculated as follows:

  nviolet=cv violetnviolet=3.00× 1081.9× 108=1.57

Refractive index corresponding to red light nredis calculated as follows:

  nred=cv rednred=3.00× 1082.00× 108=1.50

Conclusion:

Refractive index for visible light ranges from 1.50 to 1.57

To determine

(b)

A white light is incident on the medium from air, making an angle of 30.0° with the normal. Compare the angles of refraction for violet light and red light.

Expert Solution
Check Mark

Answer to Problem 49QAP

Angle of refraction for violet light= 18.54°

Angle of refraction for red light= 19.45°

Explanation of Solution

Given:

For a certain optical medium the speed of light varies from a low value of  1.9×108 m/s for violet light to a high value of  2.00×108 m/s for red light.

Formula used:

Snell's law or the law of refraction is expressed as

  n1sinθi=n2sinθr

Where

  n1:Refractive index of medium1n2:Refractive index of medium2θi:Angle of incidenceθr:Angle of refraction

  sinθr=n1sinθin2

Calculation:

As per the given problem, medium 1 is air and medium 2 is the optical medium.

  n1=1

Angle of incidence, θi=30°

Refractive index of violet light, nviolet=1.57

Substituting the values of  nviolet, n1 and θi in sinθr=n1sinθin2, we get

  sinθr(violet)=n1sinθin violetsinθr(violet)=1×sin301.57sinθr(violet)=0.318θr(violet)=sin1(0.318)θr(violet)=18.54°

Refractive index of red light, nred=1.50

Substituting the values of  nred, n1 and θi in sinθr=n1sinθin2, we get

  sinθr(red)=n1sinθin redsinθr(red)=1×sin301.50sinθr(red)=0.333θr(red)=sin1(0.333)θr(red)=19.45°

Conclusion:

Angle of refraction for violet light= 18.54°

Angle of refraction for red light= 19.45°

To determine

(c)

Repeat the previous part when the incident angle is 60.0°.

Expert Solution
Check Mark

Answer to Problem 49QAP

Explanation of Solution

Given:

For a certain optical medium the speed of light varies from a low value of  1.9×108 m/s for violet light to a high value of  2.00×108 m/s for red light.

Formulaused:

Snell's law or the law of refraction is expressed as

  n1sinθi=n2sinθr

Where

  n1:Refractive index of medium1n2:Refractive index of medium2θi:Angle of incidenceθr:Angle of refraction

  sinθr=n1sinθin2

Calculation:

As per the given problem, medium 1 is air and medium 2 is the optical medium.

  n1=1

Angle of incidence, θi=60°

Refractive index of violet light, nviolet=1.57

Substituting the values of  nviolet, n1 and θi in sinθr=n1sinθin2, we get

  sinθr(violet)=n1sinθin violetsinθr(violet)=1×sin601.57sinθr(violet)=0.551θr(violet)=sin1(0.551)θr(violet)=33.43°

Refractive index of red light, nred=1.50

Substituting the values of  nred, n1 and θi in sinθr=n1sinθin2, we get

  sinθr(red)=n1sinθin redsinθr(red)=1×sin601.50sinθr(red)=0.333θr(red)=sin1(0.577)θr(red)=35.23°

Conclusion:

Angle of refraction for violet light= 33.43°

Angle of refraction for red light= 35.23°

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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
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Laws of Refraction of Light | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=4l2thi5_84o;License: Standard YouTube License, CC-BY