Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution
Check Mark

Answer to Problem 73P

The object distance is 3.8mand 2.6m for 24mm and 36mm respectively.

Explanation of Solution

The focal length of the lens is 8.00cm and the image distances are 5.0cm, 14.0cm, 16.0cm and 20.0cm.

Write the thin lens equation

  1f=1p+1q

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

Rearrange for q

  q=pfpf                                                                                (I)

Write the equation for transverse magnification

  m=qp                                                                             (II)

Here, m is the magnification. A negative magnification implies the image if inverted.

If |m|>1, the image is enlarged, |m|<1, the image is diminished and |m|=1, the image size is same as the object size.

For the object distance of 5.0cm,

Substitute 8.00cm for f and 5.0cm for p in (I) to find q

  q=(5.0cm)(8.00cm)(5.0cm)(8.00cm)=13.3cm

Substitute 5.0cm for p and  13.3cm for q in (II) to find m

  m=(13.3cm)5.0cm=2.67

Thus, the image distance is 13.3cm, the image is virtual, upright and enlarged since q is negative, m is positive and |m|>1.

For the object distance of 14.0cm,

Substitute 8.00cm for f and 14.0cm for p in (I) to find q

  q=(14.0cm)(8.00cm)(14.0cm)(8.00cm)=18.7cm

Substitute 14.0cm for p and  18.7cm for q in (II) to find m

  m=(18.7cm)14.0cm=1.33

Thus, the image distance is 18.7cm, the image is real, inverted and enlarged since q is positive, m is negative and |m|>1.

For the object distance of 16.0cm,

Substitute 8.00cm for f and 16.0cm for p in (I) to find q

  q=(16.0cm)(8.00cm)(16.0cm)(8.00cm)=16.0cm

Substitute 16.0cm for p and  16.0cm for q in (II) to find m

  m=(16.0cm)16.0cm=1.00

Thus, the image distance is 16.0cm, the image is real, inverted and same as the object since q is positive, m is negative and |m|=1.

For the object distance of 20.0cm,

Substitute 8.00cm for f and 20.0cm for p in (I) to find q

  q=(20.0cm)(8.00cm)(20.0cm)(8.00cm)=13.3cm

Substitute 20.0cm for p and  13.3cm for q in (II) to find m

  m=(13.3cm)20.0cm=0.667

Thus, the image distance is 13.3cm, the image is real, inverted and diminished since q is positive, m is negative and |m|<1.

The results are summarized in the table:

p in cmq in cmmReal or VirtualOrientationRelative size
5.0013.32.67virtualuprightenlarged
14.018.71.33realinvertedenlarged
16.016.01.00realinvertedsame
12.013.30.667realinverteddiminished

(b)

To determine

The image height for different object distance.

(b)

Expert Solution
Check Mark

Answer to Problem 73P

The image height for 5.00cm and 20.0cm are 10.7cm and 2.67cm respectively.

Explanation of Solution

The image height is 4.00 cm.

Write the equation for transverse magnification

  m=hh

Here, h is the object height and h is the image height.

Rearrange for h

  h=mh                                                                             (III)

For the object distance of 5.0cm,

Substitute 2.67 for m and 4.00 cm for h in (III) to find h

  h=(2.67)(4.00 cm)=10.7cm

For the object distance of 20.0cm,

Substitute 0.667 for m and 4.00 cm for h in (III) to find h

  h=(0.667)(4.00 cm)=2.67cm

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Chapter 23 Solutions

Physics

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