PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 23, Problem 72P

(a)

To determine

The image distance and characteristics of the image.

(a)

Expert Solution
Check Mark

Answer to Problem 72P

The image distance and characteristics of the images is given in table 1.

Explanation of Solution

Write an expression for lens’ law.

    1f=1p+1p                                                                                                                (I)

Here, p is the object distance, q is the image distance and f is the focal length.

Rearrange equation (I) to find q.

    q=1(1f1p)                                                                                                             (II)

Write an expression for magnification.

    m=qp                                                                                                                 (III)

Here, m is the magnification.

If the magnification is positive, the image is virtual and upright. If the magnification is negative the image is real and inverted. If the magnification is greater than one, the image is larger than the object. If the magnification is less than one, the image is smaller than the object.

Conclusion:

For object distance 5.00cm:

Substitute 8.00cm for f and 5.00cm for p in equation (II) to find q.

    q=1(18.00cm15.00cm)=(8.00cm)(5.00cm)(5.00cm)(8.00cm)=3.08cm

Substitute 3.08cm for q and 5.00cm for p in equation (III) to find m.

    q=3.08cm5.00cm=3.08cm5.00cm=0.615

For object distance 8.00cm:

Substitute 8.00cm for f and 8.00cm for p in equation (II) to find q.

    q=1(18.00cm18.00cm)=(8.00cm)(8.00cm)(8.00cm)(8.00cm)=4.00cm

Substitute 4.00cm for q and 8.00cm for p in equation (III) to find m.

    q=4.00cm4.00cm=4.00cm8.00cm=0.500

For object distance 14.0cm:

Substitute 8.00cm for f and 14.0cm for p in equation (II) to find q.

    q=1(18.00cm114.0cm)=(8.00cm)(5.00cm)(14.0cm)(8.00cm)=5.09cm

Substitute 5.09cm for q and 14.0cm for p in equation (III) to find m.

    q=5.09cm14.0cm=5.09cm14.0cm=0.364

For object distance 16.0cm:

Substitute 8.00cm for f and 16.0cm for p in equation (II) to find q.

    q=1(18.00cm116.0cm)=(8.00cm)(16.0cm)(16.0cm)(8.00cm)=5.33cm

Substitute 5.33cm for q and 16.0cm for p in equation (III) to find m.

    q=5.33cm16.0cm=5.33cm16.0cm=0.333

For object distance 20.0cm:

Substitute 8.00cm for f and 20.0cm for p in equation (II) to find q.

    q=1(18.00cm120.0cm)=(8.00cm)(20.0cm)(20.0cm)(8.00cm)=5.71cm

Substitute 5.71cm for q and 20.0cm for p in equation (III) to find m.

    q=5.71cm20.0cm=5.71cm20.0cm=0.286

Thus, the image distance and characteristics of the images is given in table 1.

p(cm)q(cm)m=q/pReal or virtualOrientationRelative size
5.003.080.615VirtualUprightDiminished
8.004.000.500VirtualUprightDiminished
14.05.090.364VirtualUprightDiminished
16.05.330.333VirtualUprightDiminished
20.05.710.286VirtualUprightDiminished

Table 1

(b)

To determine

The height of the image.

(b)

Expert Solution
Check Mark

Answer to Problem 72P

  The height of the image at 5.00 cm is 2.46cm and at 20.0cm is 1.14cm.

Explanation of Solution

Write an expression for the image height.

    h'=mh                                                                                                             (IV)

Here, m is the magnification, h is the object height and h' is the image height.

Conclusion:

For object distance 5.00cm:

Substitute 4.00cm for h and 0.615 for m in equation (IV) to find h'.

    h'=(0.615)(4.00cm)=2.46cm

For object distance 20.0cm:

Substitute 4.00cm for h and 0.286 for m in equation (IV) to find h'.

    h'=(0.286)(4.00cm)=1.14cm

Thus, the height of the image at 5.00 cm is 2.46cm and at 20.0cm is 1.14cm.

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Chapter 23 Solutions

PHYSICS

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