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
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Chapter 36, Problem 36.44P

A converging lens has a focal length of 10.0 cm. Construct accurate ray diagrams for object distances of (i) 20.0 cm and (ii) 5.00 cm. (a) From your ray diagrams, determine the location of each image. (b) Is the image real or virtual? (c) Is the image upright or inverted? (d) What is the magnification of the image? (c) Compare your results with the values found algebraically. (f) Comment on difficulties in constructing the graph that could lead to differences between the graphical and algebraic answers.

(i)

Expert Solution
Check Mark
To determine

To draw: The ray diagram for the given focal lengthy of the lens and the given object distance.

Answer to Problem 36.44P

The ray diagram of the given criteria is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 36, Problem 36.44P , additional homework tip  1

Explanation of Solution

Introduction:

In a ray diagram in the case of lens or mirror the image is formed where two at least refracted or reflected rays coincide with each other.

Explanation:

Given info: The position of object is at 20.0cm and the given focal length is 10.0cm .

The ray diagram is shown in the figure below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 36, Problem 36.44P , additional homework tip  2

Figure (1)

(ii)

Expert Solution
Check Mark
To determine

To draw: The ray diagram for the given focal length of the lens and the given object distance.

Answer to Problem 36.44P

The ray diagram of the given criteria is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 36, Problem 36.44P , additional homework tip  3

Explanation of Solution

Introduction:

In a ray diagram in the case of lens or mirror the image is formed where two at least refracted or reflected rays coincide with each other.

Explanation:

Given info: The position of object is at 5.00cm and the given focal length is 10.0cm .

The ray diagram is shown in the figure below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 36, Problem 36.44P , additional homework tip  4

Figure (2)

(a)

Expert Solution
Check Mark
To determine
The location of the image the given cases.

Answer to Problem 36.44P

The image is at 20.0cm behind the lens for the case where object is 20.0cm in front of the lens, and the image is at 10cm in front of the lens when the object is at 5.0cm in front of the lens.

Explanation of Solution

From Figure (1), it is evident that the image is formed on the rear end of the lens and the image distance measured is 20.0cm .

From Figure (2), the image is formed at 10.0cm front of the lens at the focal length of the lens.

Conclusion:

Therefore, the measured distance for the image for the case when object is at 20.0cm is 20.0cm back of the mirror and for the case when object is at the 5.00cm is at 10.0cm in front of the lens.

(b)

Expert Solution
Check Mark
To determine
The image is real or virtual.

Answer to Problem 36.44P

For the case when object is at 20.0cm the image is real and for the case object is at 5.00cm the image is virtual.

Explanation of Solution

From Figure (1), it is evident that the image is formed on the rear side and is real and the images formed at the back side of the lens are real.

From Figure (2), the image is formed at 10.00cm front of the lens and the images formed on the front side of the lens are virtual.

Conclusion:

The images formed by the lens in front of it are virtual and erect and images formed on the back side are real and inverted. Hence, image formed by the object kept at. 20cm is real and the image of the object kept at 5.00cm is virtual

(c)

Expert Solution
Check Mark
To determine
The image is upright or inverse.

Answer to Problem 36.44P

For the case when object is at 20.0cm the image is inverted and for the case object is at 5.00cm the image is upright.

Explanation of Solution

From Figure (1), it is evident that the image is formed on the rear side and is real. and

the real images are always inverted

From Figure (2), the image is formed at 10.00cm front of the lens at the images formed on the front side of the lens are virtual.

The virtual images are always upright.

Conclusion:

Therefore, the images formed by the lens in front of it are virtual and erect and images formed on the back side are real and inverted. Hence, image formed by the object kept at. 20.0cm is inverted and the image of the object kept at 5.00cm is upright.

(d)

Expert Solution
Check Mark
To determine
The magnification of the image.

Answer to Problem 36.44P

For the case when object is at 20.0cm the image has magnification 1.00 and for the case object is at 5.00cm the magnification is +2.00

Explanation of Solution

From Figure (1) it is evident that the image is formed on the rear side and is real and inverted the object height is 1.00cm and the measure height of the image is 1cm .

Formula to calculate the magnification is

M=qp (1)

M is the magnification.

q is the position of the image.

p is the position of the object.

For the object at 20.0cm substitute 20.0cm for p and 20.0cm for q as the image is inverted.

M=qp=20.0cm20.0cm=1.00 (2)

For the object at the distance of 5.00cm , substitute 5.0cm for p and 10.0cm for q as the image is upright in equation (1).

M=qp=10.0cm5.00cm=+2.00 (3)

Conclusion:

Therefore, for the case of object at 20cm the magnification is 1.00 and for the case  when the object is at the distance of 5.00cm the magnification is +2.00 .

(e)

Expert Solution
Check Mark
To determine
The difference between the value of the ray diagram and the algebraic calculation.

Answer to Problem 36.44P

The value obtained from the ray diagrams for the cases are same for the values obtained in the algebraic calculation.

Explanation of Solution

Given Info: The focal length of the give lens is 10.0cm

From Figure (1) the image distance for the object at 20.0cm the image distance was +20.0cm at the back of the lens.

For algebraic calculation, the formula for the image distance is,

1f=1p+1q (4)

Here,

f is the focal length of the lens.

p is the object position.

q is the image position.

Substitute 10.0cm for f , 20.0cm for p in equation (4),

1f=1p+1q110.0cm=120cm+1q1q=120cmq=+20.0cm (5)

Form figure (1) the magnification is 1.00 as the height of image is equal to height of object.

Formula to calculate the magnification of the image

M=qp (6)

Here

M is the magnification.

Substitute 20.0cm for p and +20.0cm for q from equation (5) in equation (6)

M=qp=+20.0cm20.0cm=1.00 (7)

From figure (2) the image distance is 10.0cm in front of the mirror for the given object distance 5.00cm .

From equation (4) formula to calculate the image distance is,

1f=1p+1q

Substitute 10.0cm for f and 5.00cm for p in above equation,

1f=1p+1q110.0cm=15.00cm+1qq=10.0cm (8)

The image distance is 10.0cm in front of the mirror.

From equation (6) the formula to calculate the magnification is,

M=qp

Substitute 10.0cm for q and 5.00cm for p in the above equation.

M=qpM=(10.0cm5.00cm)M=+2.00 (9)

From equation (6) and equation (7) it is evident that for the case of object at 20cm the image distance and the magnification is same as in the case for the ray diagram, the image distance is positive hence the image is behind the lens thus, the image is real and magnification is negative hence the image is inverted.

From equation (8) and (9) the image distance and magnification is same for the ray diagram and the algebraic case when the object is 5.00cm . The image distance is negative hence a virtual image is formed and the magnification is in positive therefore the image is upright.

Conclusion:

Therefore, the result for the ray diagrams and algebraic calculations are same.

(f)

Expert Solution
Check Mark
To determine
The difficulties is making the graph which may lead to difference between the algebraic and graphical values.

Answer to Problem 36.44P

Human hand errors, Parallax errors and scale measurement errors could lead to difference in the values of graph and algebraic calculation.

Explanation of Solution

While drawing the graph the possible errors are human hand errors, parallax errors and scale measurement errors.

Human Hand Errors are while making the ray diagrams the rays might not converge with extreme precision. Parallax error/Human eye errors occur due to human eye. Scale errors are during the scale measurement.

Conclusion:

Therefore, the three most common errors that can lead to difficulties in constructing the graph which might lead to change the algebraic and graphical values are human hand errors, parallax errors and Scale errors.

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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...
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