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

The locations and sizes of the images formed by the two lenses.

Expert Solution & Answer
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Answer to Problem 5P

The location of the image formed by the first lens is 12.0cm_ and that by second lens is 4.0cm_, the height of the first image is 4.00mm, and the height of the second image is 4.0mm_.

Explanation of Solution

Given that f1 is 4.00cm, f2 is 2.00cm, the distance between the lenses is 8.00cm, the object distance from the first lens is +6.00cm,and the height of the object is 2.00mm.

Write the thin lens formula.

  1p+1q=1f                                                                                                        (I)

Here, p is the object distance from the lens, q is the image distance from the lens, f is the focal length of the lens.

Solve equation (I) for q.

  q=1(1f1p)                                                                                                      (II)

Rewrite equation (II) for the image distance of the first lens and the image distance of the second lens.

  q1=1(1f11p1)                                                                                                  (III)

  q2=1(1f21p2)                                                                                                (IV)

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

  p2=sq1                                                                                                         (V)

Write the expression for the magnification of the image in terms of object distance and image distance.

  m=qp                                                                                                            (VI)

Here, m is the magnification.

Write the expression for magnification of image in terms of height of object and image.

  m=hh                                                                                                             (VII)

Here, h is the height of the image, and h is the height of the object.

Equate the right hand sides of equations (VI) and (VII) and solve for h.

  qp=hhh=qhp                                                                                                     (VIII)

Conclusion:

Substitute 4.00cm for f1, 6.00cm for p1 in equation (III) to find q1.

    q1=1(14.00cm16.00cm)=12.0cm

Substitute 12.0cm for q1, and 8.00cm for s in equation (V) to find p2.

    p2=8.00cm12.0cm=4.0cm

Substitute 4.0cm for p2, and 2.00cm for f2 in equation (IV) to find q2.

  q2=1(12.00cm14.0cm)=4.0cm

Substitute 12.0cm for q and 2.00mm for h and 6.00cm for p in equation (VIII) to find the height of the image formed by the first lens (h1).

    h1=(12.0cm)(2.00mm)6.00cm=(12.0cm×10mm1m)(2.00mm)6.00cm×10mm1m=4.00mm

Since the image of the first lens is the object for the second lens, the height of the first lens’s image has to be taken as the height of the object of the second lens.

Substitute 4.0cm for q and 4.00mm for h and 4.00cm for p in equation (VI) to find the height of the image formed by the first lens (h2)

    h2=(4.0cm×10mm1cm)(4.00mm)4.0cm×10mm1cm=4.0mm

Therefore, the location of the image formed by the first lens is 12.0cm_ and that by second lens is 4.0cm_, the height of the first image is 4.00mm, and the height of the second image is 4.0mm_.

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Chapter 24 Solutions

Physics

Ch. 24.5 - Prob. 24.7PPCh. 24.6 - Prob. 24.6CPCh. 24.6 - Prob. 24.8PPCh. 24 - Prob. 1CQCh. 24 - Prob. 2CQCh. 24 - Prob. 3CQCh. 24 - Prob. 4CQCh. 24 - Prob. 5CQCh. 24 - Prob. 6CQCh. 24 - Prob. 7CQCh. 24 - Prob. 8CQCh. 24 - Prob. 9CQCh. 24 - 10. For human eyes, about 70% of the refraction...Ch. 24 - Prob. 11CQCh. 24 - Prob. 12CQCh. 24 - Prob. 13CQCh. 24 - Prob. 14CQCh. 24 - Prob. 15CQCh. 24 - Prob. 16CQCh. 24 - Prob. 17CQCh. 24 - Prob. 18CQCh. 24 - Prob. 19CQCh. 24 - Prob. 1MCQCh. 24 - Prob. 2MCQCh. 24 - Prob. 3MCQCh. 24 - Prob. 4MCQCh. 24 - Prob. 5MCQCh. 24 - Prob. 6MCQCh. 24 - Prob. 7MCQCh. 24 - Prob. 8MCQCh. 24 - Prob. 9MCQCh. 24 - Prob. 10MCQCh. 24 - Prob. 1PCh. 24 - Prob. 2PCh. 24 - Prob. 3PCh. 24 - Prob. 4PCh. 24 - Prob. 5PCh. 24 - Prob. 6PCh. 24 - Prob. 7PCh. 24 - Prob. 8PCh. 24 - Prob. 9PCh. 24 - Prob. 10PCh. 24 - Prob. 11PCh. 24 - Prob. 12PCh. 24 - Prob. 13PCh. 24 - Prob. 14PCh. 24 - Prob. 15PCh. 24 - Prob. 16PCh. 24 - Prob. 17PCh. 24 - Prob. 18PCh. 24 - Prob. 19PCh. 24 - Prob. 20PCh. 24 - Prob. 21PCh. 24 - Prob. 22PCh. 24 - Prob. 23PCh. 24 - Prob. 24PCh. 24 - Prob. 25PCh. 24 - Prob. 26PCh. 24 - Prob. 27PCh. 24 - Prob. 28PCh. 24 - Prob. 29PCh. 24 - Prob. 30PCh. 24 - 31. Suppose the distance from the lens to the...Ch. 24 - 32. ✦ Veronique is nearsighted; she cannot see...Ch. 24 - Prob. 33PCh. 24 - Prob. 35PCh. 24 - Prob. 37PCh. 24 - Prob. 36PCh. 24 - Prob. 38PCh. 24 - Prob. 34PCh. 24 - Prob. 39PCh. 24 - Prob. 40PCh. 24 - Prob. 41PCh. 24 - Prob. 42PCh. 24 - Prob. 43PCh. 24 - Prob. 44PCh. 24 - Prob. 45PCh. 24 - Prob. 46PCh. 24 - Prob. 47PCh. 24 - Prob. 48PCh. 24 - Prob. 49PCh. 24 - Prob. 50PCh. 24 - Prob. 51PCh. 24 - Prob. 52PCh. 24 - Prob. 53PCh. 24 - Prob. 54PCh. 24 - Prob. 55PCh. 24 - Prob. 56PCh. 24 - Prob. 57PCh. 24 - Prob. 58PCh. 24 - Prob. 59PCh. 24 - Prob. 61PCh. 24 - Prob. 60PCh. 24 - 62. ✦ The eyepiece of a Galilean telescope is a...Ch. 24 - Prob. 63PCh. 24 - Prob. 64PCh. 24 - Prob. 65PCh. 24 - Prob. 66PCh. 24 - Prob. 67PCh. 24 - Prob. 68PCh. 24 - Prob. 69PCh. 24 - Prob. 70PCh. 24 - Prob. 71PCh. 24 - Prob. 72PCh. 24 - Prob. 73PCh. 24 - Prob. 74PCh. 24 - Prob. 75PCh. 24 - Prob. 76PCh. 24 - Prob. 77PCh. 24 - Prob. 78PCh. 24 - Prob. 79PCh. 24 - Prob. 80PCh. 24 - Prob. 81PCh. 24 - Prob. 82PCh. 24 - Prob. 83PCh. 24 - Prob. 84PCh. 24 - Prob. 85PCh. 24 - 86. (a) What is the angular size of the Moon as...Ch. 24 - Prob. 86P
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