Question
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Chapter 38, Problem 79PQ

(a)

To determine

The location and magnification of the final image formed by the two lenses.

(a)

Expert Solution
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Answer to Problem 79PQ

The image is formed at a distance of 17.6cm to the right of the diverging lens. The final magnification produced by two lenses is 5.29.

Explanation of Solution

Write the expression for focal length of a thin lens.

    1f=1d0+1di1di=1f1d0=d0ffd0di=fd0d0f

Here, di is the image distance, d0 is the object distance and f is the focal length of the lens.

Write the expression for image distance of the first lens.

    di1=f1d01d01f1                                                                                              (I)

Here, di1 is the image distance of the first lens, f1 is the focal length of the first lens and d01 is the object distance of the first lens.

Write the expression for magnification by first lens.

    M1=di1d01                                                                                                   (II)

Here, M1 is the magnification produced by first lens.

Write the expression for object distance from the second lens.

    d02=ddi1                                                                                              (III)

Here, d02 is the object distance from the second lens and d is the distance between the two lenses.

Write the expression for image distance for the second lens.

    di2=f2d02d02f2                                                                                             (IV)

Here, f2 is the focal length of the second lens, di2 is the image distance from the second lens.

Write the expression for magnification by second lens.

    M2=di2d02                                                                                                (V)

Here, M2 is the magnification produced by the second lens.

Write the expression for final magnification.

    M=M1×M2                                                                                             (VI)

Conclusion:

Substitute 22cm for d01 and 15cm for f1 in equation (I) to find di1.

    di1=(15cm)(22cm)22cm15cm=47.14cm

Substitute 22cm for d01 and 47.14cm for di1 in equation (II) to find M1.

    M1=47.14cm22cm=2.14

Substitute 40cm for d and 47.14cm for di1 in equation (III) to find d02.

    d02=40cm47.14cm=7.14cm

Substitute 7.14cm for d02 and 12cm for f2 in equation (IV) to find di2    di2=(12cm)(7.14cm)7.14cm(12cm)=17.6cm

Substitute 7.14cm for d02 and 17.63cm for di2 in equation (V) to find M2.

    M2=17.63cm7.14cm=2.47

Substitute 2.14 for M1 and 2.47 for M2 in equation (VI) to find M.

    M=(2.47)(2.14)=5.29

Therefore, the image is formed at a distance of 17.6cm to the right of the diverging lens. The final magnification produced by two lenses is 5.29.

(b)

To determine

The final image formed is upright or inverted.

(b)

Expert Solution
Check Mark

Answer to Problem 79PQ

The final image formed will be inverted.

Explanation of Solution

The image formed will be real as the final image distance is positive and the image formed will be inverted as the final magnification produced by two lenses is negative. Therefore, the final image formed will be real and inverted.

Conclusion:

Therefore, the final image formed will be inverted.

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Chapter 38 Solutions

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term

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