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 10E

(a)

To determine

The location of the image graphically.

(a)

Expert Solution
Check Mark

Answer to Problem 10E

The location of the image is shown in the figure below.

Explanation of Solution

Write the expression for the focal length of the convex lens.

  1f=1v1u

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

Rearrange the above expression in terms of 1/v.

  1v=1f+1u        (I)

Conclusion:

Substitute 0.2m for f and 0.08m for u in equation (I).

  1v=10.2m10.08m=(20504)m=(304)m

Simplify the above expression in terms of v.

  v=(215)m=0.133m

The ray diagram for the required condition is drawn below.

General Physics, 2nd Edition, Chapter 24, Problem 10E

Here, the image distance from the lens is 0.133m, virtual and on the same side as the object.

Thus, the location of the image formed by the lens is shown above.

(b)

To determine

The location of the image algebraically.

(b)

Expert Solution
Check Mark

Answer to Problem 10E

The location of the image is 0.133m, virtual and on the same sideas the object.

Explanation of Solution

Write the expression for the focal length of the convex lens.

  1f=1v1u

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=1f+1u        (I)

Conclusion:

Substitute 0.2m for f and 0.08m for u in equation (I).

  1v=10.2m10.08m=(20504)m=(304)m

Simplify the above expression in terms of v.

  v=(215)m=0.133m

Thus, the location of the image is 0.133m, virtual and on the same sideas the object.

(c)

To determine

The magnification of the lens.

(c)

Expert Solution
Check Mark

Answer to Problem 10E

The magnification of the lens is +1.66.

Explanation of Solution

Write the expression for the focal length of the convex lens.

  1f=1v1u

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=1f+1u        (I)

Write the expression for the magnification of the lens.

  m=vu        (II)

Here, m is the magnification of the lens.

Conclusion:

Substitute 0.2m for f and 0.08m for u in equation (I).

  1v=10.2m10.08m=(20504)m=(304)m

Simplify the above expression in terms of v.

  v=(215)m=0.133m

Substitute 0.133m for v and 0.08m for u in equation (II).

  m=(0.133m)(0.08m)=+1.66

Thus, the magnification of the lens is +1.66.

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Chapter 24 Solutions

General Physics, 2nd Edition

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