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69 through 79 76, 78 75, 77 More lenses. Object Ostands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-8 refers to (a) the lens type, converging (C)or diverging (D), (b) the focal distance f, (c) the object distance p, (d) the image distance i, and (e) the lateral magnification m. (All distances are in centimeters.) It also refers to whether (f) the image is real (R)or virtual (V), (g) inverted(I)or non-inverted (NI) from, and (h) on the same side of the lens asO or on the opposite side. Fill in the missing information, including the value of m when only an inequality is given, where only a sign is missing, answer with the sign.

Short Answer

Expert verified
  1. The lens type is converging.
  2. The focal distance is +80cm.
  3. The object distance is +16cm.
  4. The image distance is -20cm.
  5. The lateral magnification is +1.25.
  6. The image is virtual (V).
  7. The image is non-inverted (NI).
  8. The image is on the same side of the lens as the object.

Step by step solution

01

The given data

  1. The object distance, p=+16.0cm.
  2. The lateral magnification, m=+1.25.
02

Understanding the concept of properties of the lens

An object, when placed in front of a lens, produces an image. It could be real or virtual, magnified or diminished, inverted or not inverted. The characteristics of the image are decided by the type of lens used, the focal length of the lens, and the distance of the object from the lens.

Formulae:

The lens formula, 1f=1p+1i (i)

The magnification formula of the lens, m=-ip (ii)

03

a) Calculation of the lens type

The image formed is greater than the object as the magnification m>1.0. Also the magnification is positive. Hence the image will be virtual, not inverted. This type of image can be formed only with a convergent type of lens.

Hence, the lens used here is convergent.

04

b) Calculation of the focal distance

As the lens used is convergent, the focal distance should be taken as positive. Now, using the given data in equation (i), the focal distance can be given as follows:

1f=116+1-20=180f=+80cm

Hence, the focal distance is +80cm.

05

c) Calculation of the object distance

From the given data in table, the object distance is +16cm.

06

d) Calculation of the image distance

The image distance is given using the data in equation (ii) as follows:

i=-mp

=-1.2516=-20cm

Hence, the image distance is -20cm.
07

e) Calculation of the lateral magnification 

From the given data in table, the lateral magnification is +1.25.

08

f) Calculation of the type of image

The image is virtual because the image distance is negative.

Hence, the image is virtual (V).

09

g) Calculation if the image is inverted or not

The value of lateral magnification is positive.

Hence, the image is non-inverted (NI).

10

h) Calculation of the position of the object

The image is on the same side of the lens as O since the image distance is negative.

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

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Most popular questions from this chapter

58 through 67 61 59 Lenses with given radii. Objectstands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance, index of refraction n of the lens, radiusof the nearer lens surface, and radius of the farther lens surface. (All distances are in centimetres.) Find (a) the image distanceand (b) the lateral magnificationof the object, including signs. Also, determine whether the image is (c) realor virtual, (d) invertedfrom object or non-inverted, and (e) on the same side of the lens as objector on the opposite side.

The formula 1p+1i=1f is called the Gaussian form of the thin-lens formula. Another form of this formula, the Newtonian form, is obtained by considering the distance xfrom the object to the first focal point and the distancex' from the second focal point to the image. Show thatxx'=f2 is the Newtonian form of the thin-lens formula


Isaac Newton, having convinced himself (erroneously as it turned out) that chromatic aberration is an inherent property of refracting telescopes, invented the reflecting telescope, shown schematically in Fig. 34-59. He presented his second model of this telescope, with a magnifying power of 38, to the Royal Society (of London), which still has it. In Fig. 34-59, incident light falls, closely parallel to the telescope axis, on the objective mirror. After reflection from the small mirror (the figure is not to scale), the rays form a real, inverted image in the focal plane (the plane perpendicular to the line of sight, at focal point F). This image is then viewed through an eyepiece. (a) Show that the angular magnification for the device is given by Eq. 34-15:

m=fob/fey

fob

the focal length of the objective is a mirror and

feyis that of the eyepiece.

(b) The 200 in. mirror in the reflecting telescope at Mt. Palomar in California has a focal length of 16.8 m. Estimate the size of the image formed by this mirror when the object is a meter stick 2.0 km away. Assume parallel incident rays. (c) The mirror of a different reflecting astronomical telescope has an effective radius of curvature of 10 m (鈥渆ffective鈥 because such mirrors are ground to a parabolic rather than a spherical shape, to eliminate spherical aberration defects). To give an angular magnification of 200, what must be the focal length of the eyepiece?

In Fig. 34-26, stick figure Ostands in front of a spherical mirrorthat is mounted within the boxed region;the central axis through themirror is shown. The four stick figures I1to I4suggest general locationsand orientations for the images that might be produced by themirror. (The figures are onlysketched in; neither their heightsnor their distances from the mirror are drawn to scale.) (a) Whichof the stick figures could not possibly represent images? Of thepossible images, (b) which would be due to a concave mirror, (c)which would be due to a convex mirror, (d) which would be virtual,and (e) which would involve negative magnification?

A glass sphere has radius r=-50 cmand index of refraction n1=1.6paperweight is constructed by slicing through the sphere along a plane that is 2.0 cmfrom the center of the sphere, leaving height p = h = 3.0 cm. The paperweight is placed on a table and viewed from directly above by an observer who is distance d=8.0 cmfrom the tabletop (Fig. 34-39). When viewed through the paperweight, how far away does the tabletop appear to be to the observer?

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