Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 35, Problem 50AP

(a)

To determine

The position of the objects image x as a function of the objects position x .

(a)

Expert Solution
Check Mark

Answer to Problem 50AP

The position of the objects image x as a function of the objects position x is x=102458.0x6.0x .

Explanation of Solution

Given info: The initial position of the object is 0cm , the focal length of the converging lens is 26.0cm , the position of the lens is 32.0cm and the final position of the object is 12.0cm .

The formula to calculate the focal length is,

1f=1p+1q

Here,

p is the position of the object.

q is the position of the image.

Consider the position of the image is x as the object moves and the position of the lens is 32.0cm . So, the position of the object as any instant is,

p=32cmx

Substitute 32cmx for p and 26.0cm for f in above equation to find the value of q .

126.0cm=132cmx+1qq=83226x6.0x

Thus, the position of the image for the minimum magnification is 83226x6.0x .

The formula to calculate the image position with respect to the object position is,

x=32.0+q

Substitute 83226x6.0x for q in above equation to find the value of x .

x=32.0+83226x6.0xx=102458.0x6.0x

Conclusion

Therefore, the position of the objects image x as a function of the objects position x is x=102458.0x6.0x .

(b)

To determine

The pattern of the images motion with reference to a table of values.

(b)

Expert Solution
Check Mark

Answer to Problem 50AP

The pattern of the images motion with reference to a table of values is shown below.

x Object position (cm) x Image position (cm)
0 170.7
1 193.2
2 227.0
3 283.3
4 396.0
5 734.0
6
7 618.0
8 280.0
9 167.3
10 111.0
11 77.2
12 54.7

Explanation of Solution

Given info: The initial position of the object is 0cm , the focal length of the converging lens is 26.0cm , the position of the lens is 32.0cm and the final position of the object is 12.0cm .

The formula to calculate the image distance is,

x=102458.0x6.0x

Taking the integer value of the position of the object as the integer values between 0 to 12 the corresponding values of the position of the image is provided in the table shown below.

x Object position (cm) x Image position (cm)
0 170.7
1 193.2
2 227.0
3 283.3
4 396.0
5 734.0
6
7 618.0
8 280.0
9 167.3
10 111.0
11 77.2
12 54.7

Conclusion

Therefore, the pattern of the images motion with reference to a table of values is shown below.

x Object position (cm) x Image position (cm)
0 170.7
1 193.2
2 227.0
3 283.3
4 396.0
5 734.0
6
7 618.0
8 280.0
9 167.3
10 111.0
11 77.2
12 54.7

(c)

To determine

The distance that the image moves when the object moves 12.0cm to the right.

(c)

Expert Solution
Check Mark

Answer to Problem 50AP

The image moves from infinity to beyond when the object moves 12.0cm to the right.

Explanation of Solution

Given info: The initial position of the object is 0cm , the focal length of the converging lens is 26.0cm , the position of the lens is 32.0cm and the final position of the object is 12.0cm .

The formula to calculate the image distance is,

x=102458.0x6.0x

Substitute 12 for x in above equation to find the value of x .

x=102458.0(12)6.0(12)=54.7

The image position at x at zero is 170.7 and at x equal 12 the position of the image is 54.7 . So the image moves first in the in the forward direction to infinity in the right and then jumps back to minus infinity on the left and then proceeds again in the forward direction.

Conclusion

Therefore, the image moves from infinity to beyond when the object moves 12.0cm to the right.

(d)

To determine

The direction of the image when the object moves 12.0cm to the right.

(d)

Expert Solution
Check Mark

Answer to Problem 50AP

The direction of the image movement is right but is opposite during the jump when the object moves 12.0cm to the right.

Explanation of Solution

Given info: The initial position of the object is 0cm , the focal length of the converging lens is 26.0cm , the position of the lens is 32.0cm and the final position of the object is 12.0cm .

The formula to calculate the image distance is,

x=102458.0x6.0x

The direction of the movement of the image is always right but the direction is left during the time when the image jumps to a negative infinite value form the positive infinite value. The image first moves in the positive x direction but between the value of 6cm and 7cm position of the object the image jumps in the negative infinite direction.

Conclusion

Therefore, the direction of the image movement is right but is opposite during the jump when the object moves 12.0cm to the right.

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Chapter 35 Solutions

Physics for Scientists and Engineers

Ch. 35 - Two plane mirrors stand facing each other, 3.00 m...Ch. 35 - An object is placed 50.0 cm from a concave...Ch. 35 - An object is placed 20.0 cm from a concave...Ch. 35 - An object of height 2.00 cm is placed 30.0 cm from...Ch. 35 - Why is the following situation impossible? At a...Ch. 35 - A large hall in a museum has a niche in one wall....Ch. 35 - A concave spherical mirror has a radius of...Ch. 35 - An object 10.0 cm tall is placed at the zero mark...Ch. 35 - You are training to become an opticians assistant....Ch. 35 - A certain Christmas tree ornament is a silver...Ch. 35 - Review. A ball is dropped at t = 0 from rest 3.00...Ch. 35 - You unconsciously estimate the distance to an...Ch. 35 - A convex spherical mirror has a focal length of...Ch. 35 - One end of a long glass rod (n = 1.50) is formed...Ch. 35 - Prob. 18PCh. 35 - Prob. 19PCh. 35 - Figure P35.20 (page 958) shows a curved surface...Ch. 35 - To dress up your dorm room, you have purchased a...Ch. 35 - You are working for a solar energy company. Your...Ch. 35 - An object located 32.0 cm in front of a lens forms...Ch. 35 - An objects distance from a converging lens is 5.00...Ch. 35 - A contact lens is made of plastic with an index of...Ch. 35 - A converging lens has a focal length of 10.0 cm....Ch. 35 - A converging lens has a focal length of 10.0 cm....Ch. 35 - Suppose an object has thickness dp so that it...Ch. 35 - An object is placed 10.0 cm from a diverging lens...Ch. 35 - In Figure P35.30, a thin converging lens of focal...Ch. 35 - You are working for an electronics company that...Ch. 35 - Prob. 32PCh. 35 - Two rays traveling parallel to the principal axis...Ch. 35 - Josh cannot see objects clearly beyond 25.0 cm...Ch. 35 - Figure 35.34 diagrams a cross section of a camera....Ch. 35 - The refracting telescope at the Yerkes Observatory...Ch. 35 - The distance between the eyepiece and the...Ch. 35 - What are (a) the maximum angular magnification...Ch. 35 - A patient has a near point of 45.0 cm and far...Ch. 35 - The intensity I of the light reaching the CCD in a...Ch. 35 - A certain childs near point is 10.0 cm; her far...Ch. 35 - Astronomers often take photographs with the...Ch. 35 - A simple model of the human eye ignores its lens...Ch. 35 - A real object is located at the zero end of a...Ch. 35 - The distance between an object and its upright...Ch. 35 - Prob. 46APCh. 35 - Andy decides to use an old pair of eyeglasses to...Ch. 35 - Two converging lenses having focal lengths of f1 =...Ch. 35 - Two lenses made of kinds of glass having different...Ch. 35 - Prob. 50APCh. 35 - An object is placed 12.0 cm to the left of a...Ch. 35 - An object is placed a distance p to the left of a...Ch. 35 - In a darkened room, a burning candle is placed...Ch. 35 - In many applications, it is necessary to expand or...Ch. 35 - Why is the following situation impossible?...Ch. 35 - A zoom lens system is a combination of lenses that...Ch. 35 - Consider the lensmirror arrangement shown in...Ch. 35 - A floating strawberry illusion is achieved with...
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