Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 38, Problem 76PQ

Figure P38.76 shows an object placed a distance do1 from one of two converging lenses separated by s = 1.00 m. The first lens has focal length f1 = 22.0 cm, and the second lens has focal length f2 = 45.0 cm. An image is formed by light passing through both lenses at a distance di2 = 15.0 cm to the left of the second lens. a. What is the value of do1 that will result in this image position? b. Is the final image formed by the two lenses real or virtual? c. What is the magnification of the final image? d. Is the final image upright or inverted?

Chapter 38, Problem 76PQ, Figure P38.76 shows an object placed a distance do1 from one of two converging lenses separated by s

Figure P38.76

(a)

Expert Solution
Check Mark
To determine

The value of d01.

Answer to Problem 76PQ

The value of d01 is 29.25cm.

Explanation of Solution

Write the thin lens equation for first lens.

    1f1=1d0+1di1

Here, f1 is the focal length, d01 is the object distance, and di1 is the image distance for the first lens.

Rearrange the above equation to find do1 and d02 we get,

    1d0=1f11di1d01=f1di1di1f1                                         (I)

Similarly,

    d02=f2di2di2f2                                       (II)

The image of first lens acts as object for second lens. Hence, object distance for second lens is,

    d02=sdi1                              (III)

Conclusion:

Substitute 15.0cm for di2 and 45.0cm for f2. in the equation (II) to find d02.

d02=(15.0cm)(45.0cm)15.0cm - 45.0cm=11.25cm

From equation (III), we get,

di1=sd02

Substitute 1.00m for s and 11.25cm for d02 in the above equation to find di1.

    di1=1.00m(100cm1m)11.25cm=88.75cm

Substitute 88.75cm for di1 and 22.0cm for f1 in the equation (I) to find d01.

    d01=(22.0cm)(88.75cm)88.75cm-22.0cm=29.25cm

Thus, the value of d01 is 29.25cm.

(b)

Expert Solution
Check Mark
To determine

Whether the image formed by the two lenses is real or virtual.

Answer to Problem 76PQ

The image formed by the two lenses is Virtual.

Explanation of Solution

Since, the image is formed on the left side of the second lens, the image is Virtual.

(c)

Expert Solution
Check Mark
To determine

The magnification of the final image.

Answer to Problem 76PQ

The total magnification of the final image is 4.04.

Explanation of Solution

Write the expression for the total magnification of the final image.

    m=m1m2 (IV)

Here, m1 is the magnification of the first lens and is the magnification of the second lens.

Write the expression for the magnification of the first lens.

    m1=di1d01                                              (V)

Here, m1 is the magnification, di1 is the image distance and d01 is the object distance for first lens.

Write the expression for the magnification of the second lens.

    m2=di2d02                                            (VI)

Here, m2 is the magnification, di2 is the image distance and d02 is the object distance for first lens.

Conclusion:

Substitute the equations (V) and equation (VI) in the equation (IV) to find m.

    m=(di1d01)(di2d02)

Substitute 88.75cm for di1, 29.25cm for d01, 15.0cm for di2 and 11.25cm for d02 in the above equation to find m.

    m=(88.75cm29.25cm)(15.0cm11.25cm)=4.04

Thus, the total magnification of the final image is 4.04.

(d)

Expert Solution
Check Mark
To determine

Whether the final image is upright or inverted.

Answer to Problem 76PQ

The final image produced will be inverted.

Explanation of Solution

Since, the total magnification has a negative sign, the image formed will be inverted.

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Chapter 38 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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