COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
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Chapter 23, Problem 104QAP
To determine

(a)

The three smallest angles on the either side of central maximum at which no light will reach the cat's retina and it is assumed the eyeball of cat is filled with air.

Expert Solution
Check Mark

Answer to Problem 104QAP

1st smallest angle = θ1=1.1×103rad

2nd smallest angle = θ2=2.2×103rad

3rd smallest angle = θ3=3.3×103rad

Explanation of Solution

Given:

Wavelength of light = λ=550nm=5.50×107m

Width of slit = w=0.500mm=0.5×103m

Formula used:

Following equation satisfy for the diffraction of single slit,

  wsinθ=mλ

Calculation:

Since, w is very small to λ, so, θwould be very small.

In this situation, we can apply the following approximation,

  sinθθ=mλw

Now, let's plug all values in above equation,

  θ=mλw=m(5.50× 10 7m)0.5× 10 3m=m(1.1× 10 3rad)now, when m = 1,θ1=1.1×103radwhen m = 2,θ2=2.2×103radWhen m = 3,θ3=3.3×103rad

Conclusion:

So, the three smallest angles on the either side of central maximum at which no light will reach the cat's retina and medium inside the eye ball is air are 1.1×103rad,2.2×103rad,3.3×103rad.

To determine

(b)

The three smallest angles on the either side of central maximum at which no light will reach the cat's retina and it is assumed the eyeball of cat is filled with a liquid having refractive index 1.4.

Expert Solution
Check Mark

Answer to Problem 104QAP

1st smallest angle = θ1=7.86×104rad

2nd smallest angle = θ2=1.57×103rad

3rd smallest angle = θ3=2.36×103rad

Explanation of Solution

Given:

Refractive index fluid that is in the eyeball,

  n=1.4

Wavelength of light in air = λ=550nm=5.50×107m

Width of slit = w=0.500mm=0.5×103m

Formula used:

Wavelength of light in medium is written as,

  λ'=λnλ'=wavelength of light in mediumλ=wavelength of light in airn=refractive index

Calculation:

Wavelength of light in fluid,

  λ'=λn=5.50×107m1.4=3.929×107m

Now, from above formula,

  wsinθ=mλ'sinθθ=mλ'w

Let's plug all values,

  θ=mλ'w=m(3.929× 10 7m)0.5× 10 3mθ=m(7.857×104rad)now, if m = 1,θ1=7.857×104radif m = 2,θ2=1.57×103radif m = 3,θ3=2.36×103rad

Conclusion:

Thus, the three smallest angles on either side of retina of cat's eye when eyeball is filled with fluid having refractive index n=1.4 are 7.86×104rad,1.57×103radand 2.36×103rad.

To determine

(c)

The explanation of the order of fringes due to diffraction of light by cat's eye

Expert Solution
Check Mark

Answer to Problem 104QAP

Cat cannot distinguish the location of fringe on the retina.

Explanation of Solution

From above discussion, we can see that angle of diffraction of light on either side of central bright spot of the retina is very small. So, location of bright and dark fringes would be very close to each other. Cat's eye has no ability to perceive that order of fringes. So, cat would perceive it as continuous bright light not as alternate dark and bright that occurs due to diffraction.

Conclusion:

Thus, cat would not see that alternate dark and bright fringes.

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Chapter 23 Solutions

COLLEGE PHYSICS,VOLUME 1

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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