Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 24, Problem 19P

(a)

To determine

The location at which a scree has to be placed to display an image.

(a)

Expert Solution
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Answer to Problem 19P

The location at which a screen has to be placed to display an image is 12.0cm to the right of the converging lens_. Image formed by the diverging lens is virtual and it cannot be displayed on screen.

Explanation of Solution

Given that the focal length of the converging lens is 3.00cm, the distance between the object and the converging lens is 4.00cm, the focal length of the diverging lens is 5.00cm, and the distance between the converging lens and the diverging lens is 17.0cm.

Write the thin lens formula for the first lens (converging lens)

1f1=1p1+1q1 (I)

Here, f1 is the focal length of the first lens, p1 is the distance between the object and the first lens, and q1 is the distance between the image and the first lens.

Solve equation (I) for q1.

1q1=1f11p1q1=f1p1p1f1 (II)

Write the thin lens formula for the second lens.

1f2=1p2+1q2 (III)

Here, f2 is the focal length of the second lens, p2 is the distance between the object and the second lens, and q2 is the distance between the image and the second lens.

Solve equation (III) for q2.

1q2=1f21p2q2=p2f2p2f2 (IV)

Write the expression for the object distance for the second lens.

p2=sq1 (V)

Here, s is the distance of separation between the two lenses.

Use equation (II) in (IV).

p2=sf1p1p1f1 (VI)

Conclusion:

Substitute 17.0cm for s, 3.00cm for q1, and 4.00cm for p1 in equation (VI) to find p2.

p2=17.0cm(3.00cm)(4.00cm)(4.00cm)(3.00cm)=5.0cm

Substitute 5.0cm for p2, and 5.00cm for f2 in equation (IV) to find q2.

q2=(5.0cm)(5.00cm)5.0cm(5.00cm)=2.5cm

This indicates that the image is virtual, so it cannot be displayed on the screen. On the other hand, a card can be placed in between the two lenses, at 12.0cm to the right of the converging lens to intercept the light and catch the real image formed by the first lens.

Therefore, the location at which a screen has to be placed to display an image is 12.0cm to the right of the converging lens_. Image formed by the diuverging lens is virtual and it cannot be displayed on screen

(b)

To determine

The location at which a scree has to be placed to display an image when the separation of the lenses is 10.0cm.

(b)

Expert Solution
Check Mark

Answer to Problem 19P

The location at which a screen has to be placed to display an image is 3.3cm to the right of the diverging lens_. The real image formed by the converging lens cannot be displayed on the same time since the second lens intercepts the light before this image forms.

Explanation of Solution

Given that the focal length of the converging lens is 3.00cm, the distance between the object and the converging lens is 4.00cm, the focal length of the diverging lens is 5.00cm, and the distance between the converging lens and the diverging lens is 10.0cm.

Equation (VI) gives the distance between the diverging lens and its object.

p2=sf1p1p1f1

Equation (IV) gives the distance between the diverging lens and its image.

q2=p2f2p2f2

Conclusion:

Substitute 10.0cm for s, 3.00cm for f1, and 4.00cm for p1 in equation (VI) to find p2.

p2=10.0cm(3.00cm)(4.00cm)(4.00cm)(3.00cm)=2.0cm

Substitute 2.0cm for p2, and 5.00cm for f2 in equation (IV) to find q2.

q2=(2.0cm)(5.00cm)2.0cm(5.00cm)=3.3cm

This indicates that the image is real, so it can be displayed on the screen. A screen can be placed at 3.3cm to the right of the diverging lens to display this image. The real image formed by the converging lens cannot be displayed on the same time since the second lens intercepts the light before this image forms.

Therefore, the location at which a screen has to be placed to display an image is 3.3cm to the right of the diverging lens_. The real image formed by the converging lens cannot be displayed on the same time since the second lens intercepts the light before this image forms.

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Chapter 24 Solutions

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