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9, 11, 13 Spherical mirrors. Object O stands on the central axis of a spherical mirror. For this situation, each problem in Table 34-3 gives object distanceps(centimeters), the type of mirror, and then the distance (centimeters, without proper sign) between the focal point and the mirror. Find (a) the radius of curvature r(including sign), (b) the image distance i, and (c) the lateral magnification m. Also, determine whether the image is (d) real (R)or virtual (V), (e) inverted (I)from object O or non-invertedNI, and (f) on the same side of the mirror as O or on the opposite side.

Short Answer

Expert verified

(a) The radius of curvature is r=+72cm.

(b) The image distance is i=-72cm.

(c) Lateral magnification is m=+3.0.

(d) The image is virtualV.

(e) The image is non-invertedNI.

(f) The image is on the opposite side of the object

Step by step solution

01

Step 1: Identification of the given data

f=36

The mirror is concave.

02

Determining the concept

The object distance, type of mirror, and focal length is given in the problem. First, find the radius of curvature from the focal length. Then by using the mirror formula, find the image distance. Use the formula for magnification to find the lateral magnification. Using these quantities, determine whether the image is real or virtual and inverted or non-inverted. Aslo, find the position of the image.

The formulas as follows:

r=2f

1f=1p+1i

m=-ip

Where, mis the magnification, pis the pole,fis the focal length.

03

(a) Determining the radius of curvature r

The radius of curvature:

Use the following formula to find the radius of curvature;

r=2×f

Since the mirror is concave, the focal length must be positive, i.e, role="math" localid="1663069537437" f=+36

r=2×36cm=+72cm

Therefore, the radius of curvature is,r=+72cm.

04

(b) Determining the image distance i

Image distancei:

1f=1p+1i

Rearranging fori,

i=pfp-f

Plugging the values,

i=24cm×36cm24cm-36cmi=-72cm

Therefore, the image distance is i=-72cm.

05

(c) Determining the lateral magnification

Lateral magnification:

m=-ip

Therefore,

m=--72cm24cmm=+3.0

Therefore, the lateral magnification is, m=+3.0.

06

(d) Determining whether the image is real or virtual

Whether the image is real or virtual:

Since the image distance is negative, the image is virtualV.

Therefore, the image is virtualV.

07

(e) Determining whether the image is inverted or non-inverted

Whether the image is inverted or non-inverted:

As the magnification is positive, the image is non-invertedNI.

Therefore, the image is non-invertedNI.

08

(f) Determining the position of the image

Position of image:

For spherical mirrors, real images form on the side of the mirror where the object is located and virtual images form on the opposite side. Since the image is virtual, it is formed on the opposite side as the object.

Therefore, the image is on the opposite side as the object.

The mirror formula and equation for magnification to find the required quantities can be used.

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Most popular questions from this chapter

Prove that if a plane mirror is rotated through an angle a, the reflected beam is rotated through an angle 2α. Show that this result is reasonable for α=45.

A double-convex lens is to be made of glass with an index of refraction of 1.5.One surface is to have twice the radius of curvature of the other and the focal length is to be 60mm. What is the (a) smaller and (b) larger radius?

In Fig. 34-51, a box is somewhere at the left, on the central axis of the thin converging lens. The image Imof the box produced by the plane mirror is 4.00cm “inside” the mirror. The lens–mirror separation is 10.0cm, and the focal length of the lens is 2.00cm. (a) What is the distance between the box and the lens? Light reflected by the mirror travels back through the lens, which produces a final image of the box. (b) What is the distance between the lens and that final image?

95 through 100. Three-lens systems. In Fig. 34-49, stick figure O (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to O, which is at object distance p1. Lens 2 is mounted within the middle boxed region, at distance d12 from lens 1. Lens 3 is mounted in the farthest boxed region, at distance d23 from lens 2. Each problem in Table 34-10 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 the focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i3 for the (final) image produced by lens 3 (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 3 as object O or on the opposite side.

17 through 29 22 23, 29 More mirrors. Object O stands on the central axis of a spherical or plane mirror. For this situation, each problem in Table 34-4 refers to (a) the type of mirror, (b) the focal distance f, (c) the radius of curvature r, (d) the object distance p, (e) the imagedistance i, and (f) the lateral magnification m. (All distances are in centimeters.) It also refers to whether (g) the image is real (R)or virtual localid="1662996882725" (V), (h) inverted (I)or noninverted (NI)from O, and (i) on the same side of the mirror as object O or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

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