58 through 67 61 59 Lenses with given radii. An object Ostands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance p, index of refraction n of the lens, radius r1of the nearer lens surface, and radius r2of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real (R)or virtual (V), (d) inverted (I)from the object Oor non-inverted (NI), and (e) on the same side of the lens as object Oor on the opposite side

Short Answer

Expert verified
  1. The image distance is -30cm
  2. The lateral magnification of the object is +0.86
  3. The image is virtual V.
  4. The image is not inverted (NI)from the object.
  5. The image is on the same side of the object.

Step by step solution

01

The data given

  1. The object distance, p=+35cm
  2. The index of refraction of the lens, localid="1663046948767" n=1.70
  3. The radius of the nearer lens surface, r1=+42cm
  4. The radius of the farther lens surface, r2=+33cm
02

Understanding the concept of properties of the lens

Here, we need to use the concept of image formation by the thin lens. We can use the equations 34.9 and 34.10 together to solve for the image distance. The magnification of the lens can be calculated using equation 34.7. By using the values of image distance and magnification, we can determine whether the image is real or virtual, whether it is inverted or not-inverted, and whether it is on the same side of the object or the opposite side.

Formula:

The lens formula for refraction, 1f=n-11r1-1r2 (i)

The lens formula, 1f=1P+1i(ii)

The magnification formula of the lens, m=-iP (iii)

03

(a) Calculation of the object's distance

Using the given data in equation (i), the expression of the focal length of the lens in air is given as follows:

f=r1r2n-1r2-r1=+42cm+33cm1.70-1+33cm-+42cm=-220cm

The given lens is a diverging lens because focal length is negative.

For an object in front of the lens, the object distance and image distance are related to the lens’ focal length. Thus, the image distance can be given using the data in equation (ii) as follows:

1i=1f-1Pi=PfP-f=+35cm-220cm+35cm--220cm=-30cm

Hence, the value of the image distance is -30cm

04

(b) Calculation of the lateral magnification of the object

The lateral magnification is the ratio of the object distanceto the image distance. It is given using the data in equation (iii) as follows:

m=--30cm+35cm=+0.86

Hence, the value of the lateral magnification is+0.86

.

05

(c) Calculation of the behavior of the image

If the object is inside the focal point, then it is a virtual image.The image distance is also negative.

Hence, the image is virtual.

06

(d) Calculation if the image is inverted or not

The value of magnification is positive.

Hence, the image is not inverted.

07

(e) Calculation of the position of the image

The image distance is negative.

Hence, the image is on the same side of the object.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

80 through 87 80, 87 SSM WWW 83 Two-lens systems. In Fig. 34-45, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. Each problem in Table 34-9 refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i2 for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object O or non-inverted (NI), and (e) on the same side of lens 2 as object O or on the opposite side.

A grasshopper hops to a point on the central axis of a spherical mirror. The absolute magnitude of the mirror’s focal length is 40.0cm, and the lateral magnification of the image produced by the mirror is +0.200. (a) Is the mirror convex or concave? (b) How far from the mirror is the grasshopper?

80 through 87 80, 87 SSM WWW 83 Two-lens systems. In Fig. 34-45, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. Each problem in Table 34-9 refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i2for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification Mfor the system, including signs.Also, determine whether the final image is (c) real(R) or virtual (V), (d) inverted (I)from object O or non-inverted (NI), and (e) on the same side of lens 2 as the object O or on the opposite side.

One end of a long glass rod (n=1.5)is a convex surface of radius 6.0cm.An object is located in air along the axis of the rod, at a distance of 10cmfrom the convex end (a) How far apart are the object and the image formed by the glass rod? (b) Within what range of distances from the end of the rod must the object be located in order to produce a virtual image?

In Fig. 34-60, a sand grain is 3.00cmfrom thin lens 1, on the central axis through the two symmetric lenses. The distance between focal point and lens is 4.00cmfor both lenses; the lenses are separated by 8.00cm. (a) What is the distance between lens 2 and the image it produces of the sand grain? Is that image (b) to the left or right of lens 2, (c) real or virtual, and (d) inverted relative to the sand grain or not inverted?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free