EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9781305804463
Author: Jewett
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 36, Problem 36.94CP

A zoom lens system is a combination of lenses that produces a variable magnification of a fixed object as it maintains a fixed image position. The magnification is varied by moving one or more lenses along the axis. Multiple lenses are used in practice, but the effect of zooming in on an object can be demonstrated with a simple two-lens system. An object, two converging lenses, and a screen are mounted on an optical bench. Lens 1, which is to the right of the object, has a focal length of f1 = 5.00 cm, and lens 2, which is to the right of the first lens, has a focal length of f2 = 10.0 cm. The screen is to the right of lens 2. Initially, an object is situated at a distance of 7.50 cm to the left of lens 1, and the image formed on the screen has a magnification of +1.00. (a) Find the distance between the object and the screen. (b) Both lenses are now moved along their common axis while the object and the screen maintain fixed positions until the image formed on the screen has a magnification of +3.00. Find the displacement of each lens from its initial position in part (a). (c) Can the lenses be displaced in more than one way?

(a)

Expert Solution
Check Mark
To determine
The distance between the object and the screen.

Answer to Problem 36.94CP

The distance between the object and the screen is 67.5cm .

Explanation of Solution

Given info: The focal length of the left and right lenses are f1=5.00cm and f2=10.0cm . The object distance is 7.50cm to the left of the lens 1. The magnification of the image is +1.00 .

Write the expression of thin lens equation for lens 1.

1q1=1f11p1

Here,

p1 is the object distance.

q1 is the image distance.

Substitute 5.00cm for f1 and 7.50cm for p1 in above equation.

1q1=15.00cm17.50cmq1=15cm

Write the expression for magnification.

M1=q1p1

Substitute 15cm for q1 and 7.50cm for p1 in above equation.

M1=15cm7.50cm=2

Write the expression of magnification for a combination of two lenses.

M=M1M2

Substitute 2 for M1 and 1 for M in above equation.

1=(2)M2M2=12

Write the expression to calculate the magnification for lens 2.

M2=q2p2

Substitute 12 for M2 in above equation.

12=q2p2p2=2q2 (1)

Write the expression of thin lens equation for lens 2.

1p2+1q2=1f2

Here,

p2 is the object distance from the lens 2.

q2 is the image distance from the lens 2.

Substitute 2q2 for p2 and 10.00cm for f2 in above equation.

12q2+1q2=110.00cmq2=15cm

Substitute 15cm for q2 in equation (1).

p2=2×15cm=30cm

The distance between the object and the screen is,

D=p1+q1+p2+q2

Substitute 7.5cm for p1 , 15cm for q1 , 30cm for p2 and 15cm for q2 in above equation.

D=7.5cm+15cm+30cm+15cm=67.5cm

Thus, the distance between the object and the screen is 67.5cm .

Conclusion:

Therefore, the distance between the object and the screen is 67.5cm .

(b)

Expert Solution
Check Mark
To determine
The displacement of each lens from its initial position.

Answer to Problem 36.94CP

The displacement of each lens from its initial position is 1.28cm and 17.7cm or 0.927cm and 4.44cm .

Explanation of Solution

Given info: The focal length of the left and right lenses are f1=5.00cm and f2=10.0cm . The object distance is 7.50cm to the left of the lens 1. The magnification of the image is +3.00 .

Write the expression of thin lens equation for lens 1.

1p1'+1q1'=1f1

Here,

p1' is the object distance from the lens 1.

q1' is the image distance from the lens1.

Substitute 5.00cm for f1 in above equation.

1p1'+1q1'=15.00cmq1'=5p1'p1'5

Write the expression of magnification for lens 1.

M1'=q1'p1'

Substitute 5p1'p1'5 for q1' in above equation.

M1'=5p1'p1'5p1'=5p1'5

Write the expression of magnification for the combination of lenses.

M'=M1'M2'M2'=M'M1'

Substitute 3 for M' and 5p1'5 for M1' in above equation.

M2'=35p1'5=35(p1'5)

Write the expression to calculate the magnification of lens 2.

M2'=q2'p2'

Substitute 35(p1'5) for M2' in above equation.

35(p1'5)=q2'p2'q2'=35p2'(p1'5) (2)

Write the expression of lens equation for lens 2.

1p2'+1q2'=1f2

Substitute 35p2'(p1'5) for q2' and 10cm for f2 in above equation.

1p2'+135p2'(p1'5)=110cmp2'=10(3p1'10)3(p1'5)

Substitute 10(3p1'10)3(p1'5) for p2' in equation (2).

q2'=35[10(3p1'10)3(p1'5)](p1'5)=2(3p1'10) (3)

The distance between the object and the screen is,

D=p1'+q1'+p2'+q2'

Substitute 5p1'p1'5 for q1' , 10(3p1'10)3(p1'5) for p2' , 2(3p1'10) for q2' and 67.5cm for D in above equation.

p1'+5p1'p1'5+10(3p1'10)3(p1'5)+2(3p1'10)=67.5cm[[3(p1'5)]p1'+15p1'+10(3p1'10)+2(3p1'10)[3(p1'5)]]=67.5[3(p1'5)][3(p1')2+30p1'100+18(p1')2150p1'+300202.5p1'+1012.5]=021(p1')2322.5p1'+1212.5=0

Solve the above quadratic equation to find the value of p1' .

21(p1')2322.5p1'+1212.5=0(p1'8.784)(p1'6.573)=0p1'=8.784cmp1'=6.573cm

When the value of p1' is, 8.784cm .

The displacement of object is,

p1'p1=8.784cm7.50cm=1.28cm

Substitute 8.784cm for p1' in equation (3).

q2'=2(3×8.784cm10)=32.7cm

The displacement of the image is,

q2'q2=32.7cm-15.0cm=17.7cm

When the value of p1' is, 6.573cm .

The displacement of object is,

p1'p1=6.573cm7.50cm=0.927cm

Substitute 6.573cm for p1' in equation (3).

q2'=2(3×6.573cm10)=19.44cm

The displacement of the image is,

q2'q2=19.44cm-15.0cm=4.44cm

Thus, the displacement of each lens from its initial position is 1.28cm and 17.7cm or 0.927cm and 4.44cm .

Conclusion:

Therefore, the displacement of each lens from its initial position is 1.28cm and 17.7cm or 0.927cm and 4.44cm .

(c)

Expert Solution
Check Mark
To determine
Whether the lens can be displaced by more than one way.

Answer to Problem 36.94CP

It is possible to displace the lens in more than one way.

Explanation of Solution

Given info: The focal length of the left and right lenses are f1=5.00cm and f2=10.0cm . The object distance is 7.50cm to the left of the lens 1. The magnification of the image is +1.00 .

Yes the lens can be displaced in more than one way.

The first lens can be displaced 1.28cm far from the object and the second lens can be moved by 17.7cm distance toward the object.

Another way is, the first lens can be moved 0.927cm toward the object and the second lens can be moved by 4.44cm distance toward the object.

Thus, it is possible to displace the lens in more than one way.

Conclusion:

Therefore, it is possible to displace the lens in more than one way.

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Chapter 36 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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