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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 distance ps(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 localid="1662986561416" 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-inverted (NI), and (f) on the same side of the mirror as O or on the opposite side.

Short Answer

Expert verified
  1. The radius of curvature is-28cm.
  2. The image distance is-7.7cm.
  3. Magnification ratio is0.45.
  4. The image is virtual.
  5. The image is non-inverted.
  6. The image is opposite side of mirror.

Step by step solution

01

Step 1: Given data:

The object distance,p=+17cm

The focal length of the spherical mirror,f=-14cm

The mirror is convex mirror.

02

Determining the concept:

Use equations of the radius of curvature and lateral magnification for a spherical mirror. Also, need to use equation 34.4 to calculate the distance of the image. Using these values, determine the nature of the image.If the image distance is negative, then the image is virtual and as the magnification ratio is positive, the image is upright.

Formulae:

The radius of curvature is related to the focal length of the mirror by the flowing formula.

r=2f

The spherical mirror equation is,

1f=1p+1i

The magnification equation is as below.

m=-ip

Where, mis the magnification, pis the pole,f is the focal length.

03

(a) Determining the radius of curvature:

Use the following formula to find the radius of curvature.

r=2ร—f

Since the mirror is convex, the focal length must be negative, i.e., .f=-14cm

r=-2ร—14cm=-28cm

Hence, the radius of curvature is -28cm.

04

(b) Determining the image distance i:

Write the spherical mirror equation as below.

1f=1p+1i

Rearranging it for the image distance.

i=pfp-f

Substitute known values in the above equation.

i=17cmร—-14cm17--14cm=-7.7cm

Hence, the image distance is -7.7cm.

05

(c) Determining the magnification ratio:

The magnification ratio is define as,

M=-ip=7.7cm17cm=+0.45

Hence, the magnification ratio is 0.45.

06

(d) Determining whether the image is virtual or real:

Since the image distance is negative. Therefore, the image is virtual.

07

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

As magnification is positive. Thus, the image is non-inverted.

08

(f) Determining whether the image is same side of the mirror or opposite side of the mirror:

A virtual image is formed on the opposite side of the mirror from the object. Hence, the image is opposite side of the mirror.

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Most popular questions from this chapter

In Fig. 34-32, an isotropic point source of light Sis positioned at distancedfrom a viewing screen Aand the light intensityIPat pointP(level withS) is measured. Then a plane mirrorMis placed behindSat distanced. By how much isIPmultiplied by the presence of the mirror?

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?

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 O, index of refraction n of the lens, radius of the nearer lens surface, and radius of 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 or virtual , (d) inverted from the object Oor non-inverted , and (e) on the same side of the lens as object or on the opposite side.

An object is 10.0 mm from the objective of a certain compound microscope. The lenses are 300 mm apart, and the intermediate image is 50.0 mm from the eyepiece. What overall magnification is produced by the instrument?

In Fig. 34-52, an object is placed in front of a converging lens at a distance equal to twice the focal length f1of the lens. On the other side of the lens is a concave mirror of focal lengthf2separated from the lens by a distance 2(f1+f2). Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. What are (a) the distance between the lens and that final image and (b) the overall lateral magnification M of the object? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted or non-inverted relative to the object?

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