College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 23, Problem 52AP

The object in Figure P23.52 is mid-way between the lens and the mirror, which are separated by a distance d = 25.0 cm. The magnitude of the mirror’s radius of curvature is 20.0 cm, and the lens has a focal length of –16.7 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. (b) Is the image real or virtual? (c) Is it upright or inverted? (d) What is the overall magnification of the image?

Chapter 23, Problem 52AP, The object in Figure P23.52 is mid-way between the lens and the mirror, which are separated by a

Figure P23.52

(a)

Expert Solution
Check Mark
To determine
The position of the final image.

Answer to Problem 52AP

The position of the final image is located 25.3cm behind (to the right of) the mirror.

Explanation of Solution

Given info:

The distance between the lens and mirror is 25.0cm .

The radius of curvature is 20.0cm .

The focal length is 16.7cm .

Explanation:

Formula to calculate the object distance for the mirror is,

p=d2

  • p is the object distance
  • d is distance between the lens and mirror

Substitute 25.0cm for d to find p .

p=25.0cm3=12.5cm

Therefore, the position of the object for the mirror is 12.5cm .

Formula to calculate the first image distance is,

q=1(2R1p)

  • q is the image distance
  • p is the object distance
  • R is the radius of curvature

Substitute 12.5cm for p and 20.0cm for R to find q .

q=1(220.0cm112.5cm)=+50.0cm

Therefore, the position of the first image is +50.0cm .

This image is the object for the lens. Formula to calculate the object distance for the lens is,

p'=dq

  • p' is the object distance

Substitute 25.0cm for d and +50.0cm for q to find p' .

p'=25.0cm50.0cm    =25.0cm

Therefore, the position of the object is 25.0cm .

Formula to calculate the image distance for the lens is,

q'=1(1f'1p')

  • q' is the image distance
  • p' is the object distance
  • f' is the focal length

Substitute 300m for p' and 50.0cm for f' to find q' .

q'=1(1(16.7cm)1(25.0cm))=50.3cm

Therefore, the position of the image from mirror is 50.3cm in front of (to the right of) the lens.

Formula to calculate the final position of the image is,

q''=q'+d

  • q'' is the final image distance

Substitute 25.0cm for d and 50.3 cm for q' to find q'' .

q''=50.3cm+25.0cm    =25.3cm

Thus, the position of the final image is located 25.3cm behind (to the right of) the mirror.

Conclusion:

The position of the final image is located 25.3cm behind (to the right of) the mirror.

(b)

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

Answer to Problem 52AP

The final image is virtual.

Explanation of Solution

The final image position is less than zero. Thus, the final image is virtual.

Conclusion:

The final image is virtual.

(c)

Expert Solution
Check Mark
To determine
The total magnification of the system.

Answer to Problem 52AP

The total magnification of the system is +8.05 .

Explanation of Solution

Given info:

Explanation:

Formula to calculate the total magnification of the system is,

M=M1M2=(qp)(q'p')

  • M is the magnification of the total system
  • M1 is the magnification of first image
  • M2 is the magnification of second image

Substitute 50.0cm for q ,   12.5cm for p , 50.3cm for q' and 25.0cm for p' to find M .

M=(50.0cm12.5cm)(50.3cm25.0cm)=+8.05

Thus, the total magnification of the system is +8.05 .

Conclusion:

The total magnification of the system is +8.05 .

(d)

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

Answer to Problem 52AP

The final image is upright.

Explanation of Solution

The magnification is greater than zero. Thus, the final image is upright.

Conclusion:

The final image is upright.

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Chapter 23 Solutions

College Physics

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