PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 24, Problem 80P

(a)

To determine

The distances of the intermediate and final image relative to the corresponding lenses.

(a)

Expert Solution
Check Mark

Answer to Problem 80P

The distances of the intermediate and final image relative to the corresponding lenses are 1.9cm_ to the left of the first lens and 9.07cm_ to the left of the second lens.

Explanation of Solution

Write the expression for the lens maker’s formula.

    1p1+1q1=1f1

Here, p1 is the object distance, q1 is the image distance and f1 is the focal length of the first lens.

Rewrite the above equation to find the image distance of the first lens.

    q1=f1p1p1f1                                                                                                              (I)

Write the expression to find the object distance of the second lens.

    p2=sq1                                                                                                                (II)

Here, s is the separation distance of the lenses.

Write the expression for the lens maker’s formula.

    1p2+1q2=1f2

Here, p1 is the object distance, q2 is the image distance and f2 is the focal length of the second lens.

Rewrite the above equation to find the image distance of the first lens.

    q1=f1p1p1f1                                                                                                            (III)

Conclusion:

Substitute 30.0cm for f1 and 1.8cm for p1 in the equation (I).

    q1=(30.0cm)(1.8cm)1.8cm30.0cm=1.9cm

Substitute 21.0cm for s and 1.9cm for q1 in the equation (II).

    p2=21.0cm(1.9cm)=22.9cm

Substitute 15.0cm for f2 and 22.915cm for p2 in the equation (III).

    q2=(15.0cm)(22.915cm)22.915cm(15.0cm)=9.07cm

Therefore, the distances of the intermediate and final image relative to the corresponding lenses are 1.9cm_ to the left of the first lens and 9.07cm_ to the left of the second lens.

(b)

To determine

The total magnification.

(b)

Expert Solution
Check Mark

Answer to Problem 80P

The total magnification is 0.42_.

Explanation of Solution

Write the expression to find the magnification of one lens.

    m=qp

Here, m is the magnification of the lens.

Write the expression for total magnification.

    M=m1×m2

Here, m1 is the magnification of the first lens and m2 is the magnification of the magnification of the second lens.

Rewrite the above equation in terms of object and image distances.

    m=q1p1×q2p2

Conclusion:

Substitute 1.9cm for q1, 9.07cm for q2, 1.8cm for p1 and 22.9cm for p2 in the above equation.

    m=(1.9cm)1.8cm×(9.07cm)22.9cm=0.42

Therefore, the total magnification is 0.42_.

(c)

To determine

The height of the final image.

(c)

Expert Solution
Check Mark

Answer to Problem 80P

The height of the final image is 0.84mm_.

Explanation of Solution

Write the expression to find the height of the final image.

    h=mh

Here, h is the height of the initial image.

Conclusion:

Substitute 0.42 for m and 2.00mm for h in the above equation.

    h=(0.42)(2.00mm)=0.84mm

Therefore, the height of the final image is 0.84mm_.

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Chapter 24 Solutions

PHYSICS

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