Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
Question
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Chapter 38, Problem 39PQ
To determine

The distance between the image and the surface 2 and the magnification of the final image.

Expert Solution & Answer
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Answer to Problem 39PQ

The distance of the screen from the surface 2 is 7.5cm and the magnification of the final image is 0.13.

Explanation of Solution

Write the expression for the refraction at a spherical surface.

    nido+ntdi=(ntni)r                                                                           (I)

Here, ni is the index of refraction of the travelling medium, nt is the index of refraction of the denser medium, do is the object distance, dt is the image distance and r is the radius of curvature.

Rearrange the equation (I) for di.

    ntdi=(ntni)rnidodi=nt(ntni)rnido                                                                          (II)

Write the expression for the magnification for a refractive surface.

    m=nidintdo                                                                                  (III)

Here, m is the magnification for a refractive surface.

Write the expression for the total magnification for a thick lens.

    M=m1m2                                                                                          (IV)

Here, M is the total magnification, m1 is the magnification at surface 1 and m2 is the magnification at surface 2.

Conclusion:

Substitute 1.333 for nt, 1 for ni, 6.0cm for r and 100.0cm for do in equation (II) to find di.

    di=1.333(1.3331)6.0cm1100.0cm=1.3330.05550.01cm=29.3cm

The positive sign indicates that the image is formed on the right side of surface 1.

Substitute 1.333 for nt, 1 for ni, 100.0cm for do and 29.3cm for di in equation (III) to find m1.

    m1=(1)(29.3cm)(1.333)(100.0cm)=0.22

The image formed by surface 1 acts as the object for the surface 2. Thus, the object distance for surface 2 is,

    do=12.0cm29.3cm=17.3cm

Substitute 1.333 for ni, 1 for nt, 6.0cm for r and 17.3cm for do in equation (II) to find di.

    di=1(11.333)6.0cm1.33317.3cm=10.0555+0.077cm=7.5cm

The positive sign indicates that the image is formed behind the surface 2.

Substitute 1.333 for nt, 1 for ni, 17.3cm for do and 7.5cm for di in equation (III) to find m2.

    m2=(1.333)(7.5cm)(1)(17.3cm)=0.58

Substitute 0.22 for m1 and 0.58 for m2 in equation (IV) to find M.

    M=(0.22)(0.58)=0.13

Therefore, the distance of the screen from the surface 2 is 7.5cm and the magnification of the final image is 0.13.

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Chapter 38 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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