PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 24, Problem 78P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 78P

The distances of the intermediate and final image relative to the corresponding lenses are 60.0cm_ to the right of the first lens and 5.45cm_ to the right 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 second lens.

    q2=f2p2p2f2                                                                                                          (III)

Conclusion:

Substitute 15.0cm for f1 and 20.0cm for p1 in the equation (I).

    q1=(15.0cm)(20.0cm)20.0cm15.0cm=60.0cm

Substitute 50.0cm for s and 60.0cm for q1 in the equation (II).

    p2=50.0cm(60.0cm)=10.0cm

Substitute 12.0cm for f2 and 10.0cm for p2 in the equation (III).

    q2=(12.0cm)(10.0cm)10.0cm(12.0cm)=5.45cm

Therefore, the distances of the intermediate and final image relative to the corresponding lenses are 60.0cm_ to the right of the first lens and 5.45cm_ to the right of the second lens.

(b)

To determine

The total magnification.

(b)

Expert Solution
Check Mark

Answer to Problem 78P

The total magnification is 1.64_.

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 60.0cm for q1, 20.0cm for q2, 5.455cm for p1 and 10.0cm for p2 in the above equation.

    m=60.0cm20.0cm×(5.455cm)10.0cm=1.64

Therefore, the total magnification is 1.64_.

(c)

To determine

The height of the final image.

(c)

Expert Solution
Check Mark

Answer to Problem 78P

The height of the final image is 4.91cm_.

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 1.636 for m and 3.00cm for h in the above equation.

    h=(1.636)(3.00cm)=4.91cm

Therefore, the height of the final image is 4.91cm_.

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Chapter 24 Solutions

PHYSICS

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