Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
Question
Book Icon
Chapter 36, Problem 36.36P

(a)

To determine

The image position of the fishes that are located at 5.00cm and 25.0cm in front of the aquarium wall.

(a)

Expert Solution
Check Mark

Answer to Problem 36.36P

The image is 3.77cm from the front of the wall when the fish is at 5.00cm and when the fish is at 25.0cm the image is 19.3cm form the front of the wall

Explanation of Solution

Given info: The radius of curvature of the curved plastic is 2.25m . The refractive index of water is 1.33 and that of air is 1.00 .

The formula to calculate image position of the fish inside the aquarium is,

n1p+n2q=n2n1R (1)

Here,

n1 is the refractive index of the water.

n2 is the refractive index of the air.

p is the object distance.

q is the image distance.

R is the radius of curvature

The radius of curvature will be negative the centre of curvature lies on the object side.

For part (i): when the fish is at 5.00cm .

n1p+n2q=n2n1R

Substitute  1.33 for n1 , 1.00 for n2 and 2.25m for R and 25.0cm for p . in the equation (1),

n1p+n2q=n2n1R1.335.00cm+1.00q=1.0001.3332.25m1.335.00cm+1.000q=0.33332.25m×1cm102mq=3.77cm (2)

The image position for the fish at 5.00cm is 3.77cm from the front of the wall in the water.

For part (ii) when the fish is at 25.0cm .

From equation (2) the image position is,

n1p+n2q=n2n1R1.3325.0cm+1.00q=1.0001.3332.25m1.3325.0cm+1.000q=0.33332.25m×1cm102mq=19.3cm (3)

The image is at 19.3cm for the fish at 25.0cm from the front of the wall.

Conclusion:

Therefore, the image is 3.77cm from the front of the wall when the fish is at 5.00cm and when the fish is at 25.0cm the image is 19.3cm form the front of the wall.

(b)

To determine

The magnification of the images for part (a)

(b)

Expert Solution
Check Mark

Answer to Problem 36.36P

The magnification of the image is +1.01 when the fish is at 5.00cm and when the fish is at 25.0cm the magnification of the image is +1.03 .

Explanation of Solution

Explanation

Given info: The radius of curvature of the curved plastic is 2.25m . The refractive index of water is 1.33 and that of air is 1.00 .

The formula to calculate the magnification of the image is,

M=n1qn2p (4)

For part (i): when the fish is at 5.00cm .

Substitute 1.33 for n2 , 1.00 for n1 , 5.00cm for p and 3.77cm for q in equation (4).

M=n1qn2p=(1.333(3.77cm)1.00(5.00cm))=+1.01

Thus when the fish is at 5.00cm the magnification of the image is +1.01 .

For part (ii): when the fish is at 25.0cm .

Substitute 1.33 for n2 , 1.00 for n1 , 25.0cm for p and 19.3cm for q in equation(4).

M=n1qn2p=(1.333(19.3cm)1.00(25.0cm))=+1.03

Thus when the fish is at 25.0cm the magnification of the image is +1.03 .

Conclusion:

Therefore, when the fish is at 5.00cm the magnification of the image is +1.01 and when the fish is at 25.0cm the magnification of the image is +1.03 .

(c)

To determine

The reason refractive index of the plastic is not required to solve the problem.

(c)

Expert Solution
Check Mark

Answer to Problem 36.36P

The refractive index of plastic is not playing any major role in light propagation

Explanation of Solution

Explanation

The plastic has uniform thickness and the surface from which the ray is entering and the surface from which is leaving are parallel to each other. The ray might get slightly displaced, but it will not change the direction of its propagation by going through plastic air interface. The only difference will be due to water-air interface.

Conclusion:

Therefore, the ray might get slightly displaced, but it will not change the direction of its propagation by going through plastic air interface. So the refractive index of plastic is not playing any major role in light propagation.

(e)

To determine

The image distance of the fish is greater than the fish itself and the magnification

(e)

Expert Solution
Check Mark

Answer to Problem 36.36P

Yes, the image distance of the fish can be larger than the distance where the fish is itself.

Explanation of Solution

Explanation

For the object distance greater than the radius of curvature the image distance will greater than the distance at which fish is itself. If the aquarium were very long the radius of curvature will not increase therefore if the object distance is more than the radius of curvature the image of the fish will be at even farther distance away from the fish itself.

Conclusion:

Therefore, If the fish is present at distance larger than the radius of curvature the image of the fish would be even farther than that of fish itself.

(d)

To determine

The magnification of the image when the image of the fish is even farther than the position of fish itself.

(d)

Expert Solution
Check Mark

Answer to Problem 36.36P

.

Explanation of Solution

Let the object distance is greater than the radius of curvature.

For the condition

p=|R|

Formula to calculate the image distance from Lens formula

n1p+n2q=n1n2|R|

Substitute |R| for p in the above equation

n1p+n2q=n1n2|R|n1|R|+n2q=n1n2|R|n2q=n2|R|q=|R|

For the condition

p>|R|

Take reciprocal of the above question

p>|R|1R>1p1R1p>0 (5)

Formula to calculate the image distance from Lens formula,

n1p+n2q=n1n2|R|

n1p+n2q=n1n2|R|n2q=n1n2Rn1pn2q=n1Rn2Rn1p

Divide by n2 both sides the equation.

1q=1|R|+n1n2(1|R|1p)

The condition is when the image distance is greater than the radius of curvature take the magnitude of the equation.

1|q|=1|R|n1n2(1|R|1p) (6)

Substitute 1.33 for n1 and 1.00 for n2 in equation (6).

1|q|=1|R|n1n2(1|R|1p)1|q|=1|R|1.331.00(1|R|1p)1|q|<1|R| (7)

The reciprocal of the equation is

|q|>|R|

Thus the image of the fish will also be at greater distance than that of radius of curvature.

An example for the above case is let the fish is at twice the distance of the magnitude of radius of curvature.

The image of the fish is calculated from the formula from equation (7).

1|q|=1|R|1.331.00(1|R|1p)

Substitute 2.00|R| for p

1|q|=1|R|1.331.00(1|R|1p)1|q|=1|R|1.331.00(1|R|1|2R|)q=3.00|R|

Thus the image of the fish is 3|R| when the object is at 2|R| , Hence for the obj4ect at greater than the radius of curvature the image is farther than the object itself.

The formula to calculate the magnification of the image is,

M=n1qn2p

Substitute  1.33 for n1 and 1.00 for n2 2|R| for p and 3|R| for q

M=n1qn2pM=1.33(3.00|R|)1.00(|R|)=+2.00

Thus the magnification of the fish image is +2.00 if the fish is present at the distance of 2|R| and the image formed is at 3|R| .

Conclusion:

Therefore, if the fish is present at distance larger than the radius of curvature the image of the fish would be even farther than that of fish itself.

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!

Chapter 36 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 36 - An object is located 50.0 cm from a converging...Ch. 36 - Prob. 36.4OQCh. 36 - A converging lens in a vertical plane receives...Ch. 36 - Prob. 36.6OQCh. 36 - Prob. 36.7OQCh. 36 - Prob. 36.8OQCh. 36 - A person spearfishing from a boat sees a...Ch. 36 - Prob. 36.10OQCh. 36 - A converging lens made of crown glass has a focal...Ch. 36 - A converging lens of focal length 8 cm forms a...Ch. 36 - Prob. 36.13OQCh. 36 - An object, represented by a gray arrow, is placed...Ch. 36 - Prob. 36.1CQCh. 36 - Prob. 36.2CQCh. 36 - Why do some emergency vehicles have the symbol...Ch. 36 - Prob. 36.4CQCh. 36 - Prob. 36.5CQCh. 36 - Explain why a fish in a spherical goldfish bowl...Ch. 36 - Prob. 36.7CQCh. 36 - Lenses used in eyeglasses, whether converging or...Ch. 36 - Suppose you want to use a converging lens to...Ch. 36 - Consider a spherical concave mirror with the...Ch. 36 - In Figures CQ36.11a and CQ36.11b, which glasses...Ch. 36 - Prob. 36.12CQCh. 36 - Prob. 36.13CQCh. 36 - Prob. 36.14CQCh. 36 - Prob. 36.15CQCh. 36 - Prob. 36.16CQCh. 36 - Prob. 36.17CQCh. 36 - Determine the minimum height of a vertical flat...Ch. 36 - In a choir practice room, two parallel walls are...Ch. 36 - (a) Does your bathroom mirror show you older or...Ch. 36 - Prob. 36.4PCh. 36 - A periscope (Fig. P35.3) is useful for viewing...Ch. 36 - Two flat mirrors have their reflecting surfaces...Ch. 36 - Two plane mirrors stand facing each other, 3.00 m...Ch. 36 - An object is placed 50.0 cm from a concave...Ch. 36 - A concave spherical mirror has a radius of...Ch. 36 - An object is placed 20.0 cm from a concave...Ch. 36 - A convex spherical mirror has a radius of...Ch. 36 - Prob. 36.12PCh. 36 - An object of height 2.00 cm is placed 30.0 cm from...Ch. 36 - A dentist uses a spherical mirror to examine a...Ch. 36 - A large hall in a museum has a niche in one wall....Ch. 36 - Why is the following situation impossible? At a...Ch. 36 - Prob. 36.17PCh. 36 - A certain Christmas tree ornament is a silver...Ch. 36 - (a) A concave spherical mirror forms an inverted...Ch. 36 - (a) A concave spherical mirror forms ail inverted...Ch. 36 - An object 10.0 cm tall is placed at the zero mark...Ch. 36 - A concave spherical mirror has a radius of...Ch. 36 - A dedicated sports car enthusiast polishes the...Ch. 36 - A convex spherical mirror has a focal length of...Ch. 36 - A spherical mirror is to be used to form an image...Ch. 36 - Review. A ball is dropped at t = 0 from rest 3.00...Ch. 36 - You unconsciously estimate the distance to an...Ch. 36 - Prob. 36.28PCh. 36 - One end of a long glass rod (n = 1.50) is formed...Ch. 36 - A cubical block of ice 50.0 cm on a side is placed...Ch. 36 - Prob. 36.31PCh. 36 - Prob. 36.32PCh. 36 - A flint glass, plate rests on the bottom of an...Ch. 36 - Figure P35.20 (page 958) shows a curved surface...Ch. 36 - Prob. 36.35PCh. 36 - Prob. 36.36PCh. 36 - A goldfish is swimming at 2.00 cm/s toward the...Ch. 36 - A thin lens has a focal length of 25.0 cm. Locate...Ch. 36 - An object located 32.0 cm in front of a lens forms...Ch. 36 - An object is located 20.0 cm to the left of a...Ch. 36 - The projection lens in a certain slide projector...Ch. 36 - An objects distance from a converging lens is 5.00...Ch. 36 - A contact lens is made of plastic with an index of...Ch. 36 - A converging lens has a focal length of 10.0 cm....Ch. 36 - A converging lens has a focal length of 10.0 cm....Ch. 36 - A diverging lens has a focal length of magnitude...Ch. 36 - Prob. 36.47PCh. 36 - Suppose an object has thickness dp so that it...Ch. 36 - The left face of a biconvex lens has a radius of...Ch. 36 - In Figure P35.30, a thin converging lens of focal...Ch. 36 - An antelope is at a distance of 20.0 m from a...Ch. 36 - Prob. 36.52PCh. 36 - A 1.00-cm-high object is placed 4.00 cm to the...Ch. 36 - The magnitudes of the radii of curvature are 32.5...Ch. 36 - Two rays traveling parallel to the principal axis...Ch. 36 - Prob. 36.56PCh. 36 - Figure 35.34 diagrams a cross section of a camera....Ch. 36 - Josh cannot see objects clearly beyond 25.0 cm...Ch. 36 - Prob. 36.59PCh. 36 - A person sees clearly wearing eyeglasses that have...Ch. 36 - Prob. 36.61PCh. 36 - A certain childs near point is 10.0 cm; her far...Ch. 36 - A person is to be fitted with bifocals. She can...Ch. 36 - A simple model of the human eye ignores its lens...Ch. 36 - A patient has a near point of 45.0 cm and far...Ch. 36 - A lens that has a focal length of 5.00 cm is used...Ch. 36 - The distance between the eyepiece and the...Ch. 36 - The refracting telescope at the Yerkes Observatory...Ch. 36 - A certain telescope has an objective mirror with...Ch. 36 - Astronomers often take photographs with the...Ch. 36 - Prob. 36.71APCh. 36 - A real object is located at the zero end of a...Ch. 36 - The distance between an object and its upright...Ch. 36 - Prob. 36.74APCh. 36 - Andy decides to use an old pair of eyeglasses to...Ch. 36 - Prob. 36.76APCh. 36 - The lens and mirror in Figure P36.77 are separated...Ch. 36 - Two converging lenses having focal lengths of f1 =...Ch. 36 - Figure P36.79 shows a piece of glass with index of...Ch. 36 - Prob. 36.80APCh. 36 - The object in Figure P36.81 is midway between the...Ch. 36 - In many applications, it is necessary to expand or...Ch. 36 - Prob. 36.83APCh. 36 - Prob. 36.84APCh. 36 - Two lenses made of kinds of glass having different...Ch. 36 - Why is the following situation impossible?...Ch. 36 - An object is placed 12.0 cm to the left of a...Ch. 36 - An object is placed a distance p to the left of a...Ch. 36 - An observer to the right of the mirror-lens...Ch. 36 - In a darkened room, a burning candle is placed...Ch. 36 - Prob. 36.91APCh. 36 - An object 2.00 cm high is placed 40.0 cm to the...Ch. 36 - Assume the intensity of sunlight is 1.00 kW/m2 at...Ch. 36 - A zoom lens system is a combination of lenses that...Ch. 36 - Figure P36.95 shows a thin converging lens for...Ch. 36 - A floating strawberry illusion is achieved with...Ch. 36 - Consider the lensmirror arrangement shown in...
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
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
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, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning