General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 24, Problem 15E

(a)

To determine

The lens-to-film distance.

(a)

Expert Solution
Check Mark

Answer to Problem 15E

The lens-to-film distance is 0.0508m.

Explanation of Solution

Write the expression for the focal length of the lens.

  1f=1v+1u                                                                                        

Here, f is the focal length of the lens, v is the image distance and u is the object distance.

Rearrange the above expression for the image distance.

  1v=1f1u        (I)

Conclusion:

Substitute 0.05m for f and 3m for u in equation (I).

  1v=1(0.05m)1(3m)=(2013)m=593m

Simplify the above expression for the image distance.

  v=(359)m=0.0508m

Thus, the lens-to-film distance is 0.0508m.

(b)

To determine

The linear magnification of the lens.

(b)

Expert Solution
Check Mark

Answer to Problem 15E

The linear magnification of the lens is 0.0169.

Explanation of Solution

Write the expression for the focal length of the lens.

  1f=1v+1u                                                                                        

Here, f is the focal length of the lens, v is the image distance and u is the object distance.

Rearrange the above expression for the image distance.

  1v=1f1u        (I)

Write the expression for the magnification of the lens.

  m=vu        (II)

Here, m is the magnification of the lens.

Conclusion:

Substitute 0.05m for f and 3m for u in equation (I).

  1v=1(0.05m)1(3m)=(2013)m=593m

Simplify the above expression for the image distance.

  v=(359)m=0.0508m

Substitute 0.0508m for v and 3m for u in equation (i).

  m=(0.0508m)(3m)=0.0169

Thus, the linear magnification of the lens is 0.0169.

(c)

To determine

The maximum height of the person.

(c)

Expert Solution
Check Mark

Answer to Problem 15E

The maximum height of the person is 1.42m.

Explanation of Solution

Write the expression for the focal length of the lens.

  1f=1v+1u                                                                                        

Here, f is the focal length of the lens, v is the image distance and u is the object distance.

Rearrange the above expression for the image distance.

  1v=1f1u        (I)

Write the expression for the magnification of the lens.

  m=vu        (II)

Here, m is the magnification of the lens.

Write the expression for the magnification of the lens.

  m=IO        (III)

Here, I is the image height and O is the object height.

Conclusion:

Substitute 0.05m for f and 3m for u in equation (i).

  1v=10.05m13m=(2013)m=593m

Simplify the above expression for the image distance.

  v=(359)m=0.0508m

Substitute 0.0508m for v and 3m for u in equation (ii).

  m=0.0508m3m=0.0169

Substitute 0.0169 for m and 0.024m for O in equation (iii).

  0.0169=0.024mO

Simplify the above expression for the object height.

  O=0.024m0.0169=1.42m

Thus, the maximum height of the person is 1.42m.

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Chapter 24 Solutions

General Physics, 2nd Edition

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