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Chapter 26, Problem 57P

(a)

To determine

Position of the image.

(a)

Expert Solution
Check Mark

Answer to Problem 57P

The image is 160cm_ to the left of the lens.

Explanation of Solution

Write the mirror equation for the first pass through the lens.

      1f1=1p1+1q1        (I)

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

Rewrite (I) in terms of q1.

  q1=p1f1p1f1        (II)

The object of the mirror would be at distance,

  p2=q11m        (III)

Therefore, the image distance for this case would be

    q2=p2f2p2f2        (IV)

The image formed in this mirror would be

    p3=1m+|q2|        (V)

Write the equation for image position after the second pass through the lens.

    q3=p3f3p3f3        (VI)

Conclusion:

Substitute 80cm for f1 and 100cm for p1 in (II)

    q1=(100cm)(80cm)(100cm)(80cm)=400cm=4m

Substitute 4m for q1 in (III)

    p2=4m1m=3m=300cm

Substitute 300cm for p2 and 50cm for f2 in (IV)

    q2=(300cm)(50cm)(300cm)(50cm)=60cm=0.6m

Substitute 0.6m for q2 in (V)

    p3=1m+|0.6m|=1.6m=160cm

Substitute 160cm for p3 and 80cm for f3 in (VI)

    q2=(160cm)(80cm)(160cm)(80cm)=160cm

The image is 160cm_ to the left of the lens.

(b)

To determine

Overall magnification of the image.

(b)

Expert Solution
Check Mark

Answer to Problem 57P

Overall magnification is 0.800_.

Explanation of Solution

Write the equation for magnification of three cases

    M1=q1p1M2=q2p2M3=q3p3        (VII)

Write the equation for overall magnification

    M=M1M2M3        (VIII)

Substitute (VII) in (VIII)

    M=q1q2q3p1p2p3        (IX)

Conclusion:

Substitute 400cm for q1, 60cm for q2, 160cm for q3, 100cm for p1, 300cm for p2 and 160cm for p3.

    M=(400cm)(60cm)(160cm)(100cm)(300cm)(160cm)=0.800

Overall magnification is 0.800_.

(c)

To determine

Whether the final image is upright or inverted.

(c)

Expert Solution
Check Mark

Answer to Problem 57P

The final image is inverted.

Explanation of Solution

The sign of magnification determines whether the image is upright or inverted.

If magnification is positive, the image is upright. The negative sign of magnification indicates that the image is inverted.

Conclusion:

The final image is inverted.

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Chapter 26 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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