Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 32, Problem 50P

(a)

To determine

The position of the final image.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The focal length of a converging lens is 10cm and focal length of a diverging lenses is (15cm) .

The separation between the two lenses is 35cm .

The distance of the object from the converging lens is 20cm .

Formula used:

Draw a ray diagram to show the final image position and size.

  Physics for Scientists and Engineers, Chapter 32, Problem 50P

Write the expression for thin lens equation for first image.

  1u1+1v1=1f1 …… (1)

Here, u1 is the distance of the object from converging lens, v1 is the distance of the first image and f1 is the focal length of the converging lens.

Write the expression for thin lens equation for second image.

  1u2+1v2=1f2 …… (2)

Here, u2 is the distance of the object from diverging lens, v2 is the distance of the second image and f2 is the focal length of the diverging lens.

Write the expression for the lateral magnification of image due to converging lens.

  m1=v1u1 …… (3)

Here, m1 is the lateral magnification due to converging lens.

Write the expression for the lateral magnification due to the diverging lens.

  m2=v2u2 …… (4)

Here, m2 is the lateral magnification due to the diverging lens.

Calculation:

Rewrite equation (1) to calculate the first image distance.

  v1=f1u1u1f1

Substitute 10cm for f1 and 20cm for u1 in the above equation.

  v1=( 10cm)( 20cm)( 20cm)( 10cm)v1=20cm

Substitute 20cm for u1 and 20cm for v1 in equation (3).

  m1=20cm20cmm1=1.0

The distance of the object from diverging lens is calculated below.

  u2=dv1

Here, d is the separation between the two lenses.

Substitute 20cm for v1 and 35cm for d in the above equation.

  u2=35cm20cmu2=15cm

Rewrite equation (2) to calculate the second image distance.

  v2=f2u2u2f2

Substitute (15cm) for f2 and 15cm for u2 in the above equation.

  v2=( 15cm)( 15cm)( 15cm)( 15cm)v2=7.5cm

Substitute 15cm for u2 and (7.5cm) for v2 in equation (4).

  m2=7.5cm15cmm2=0.50

The final distance of the image is calculated below.

  v=v2+d+u1

Here, v is the distance of the final image.

Substitute (7.5cm) for v2 , 20cm for u1 and 35cm for d in the above equation.

  v=(7.5cm)+35cm+20cmv48cm

Conclusion:

Thus, the distance of the final image is 48cm to the right of the object.

(b)

To determine

Whether the final image is real or virtual and upright or inverted.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The distance of the second image is (7.5cm) .

The lateral magnification of the image due to the converging lens is (1.0) .

The lateral magnification of the image due to the diverging lens is 0.50 .

Formula used:

Write the expression for the overall lateral magnification for the system of two lenses.

  m=m1m2 …… (5)

Here, is the overall lateral magnification for the system of two lenses.

Calculation:

Substitute (1.0) for m1 and 0.50 for m2 in equation (5).

  m=(1.0)(0.50)m=0.50

Conclusion:

Thus, the image is virtual and inverted. Because the image distance, u2<0 , so the image is virtual and the overall magnification, m<0 , so the image is inverted and half of the size of the object.

(c)

To determine

The overall magnification of the system.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The lateral magnification of the image due to the converging lens is (1.0) .

The lateral magnification of the image due to the diverging lens is 0.50 .

Formula used:

Write the expression for the overall lateral magnification for the system of two lenses.

  m=m1m2

Calculation:

Substitute (1.0) for m1 and 0.50 for m2 in equation (5).

  m=(1.0)(0.50)m=0.50

Conclusion:

Thus, the overall magnification for the system of two lenses is (0.50) .

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Chapter 32 Solutions

Physics for Scientists and Engineers

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