Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Through 29 22 23, 29 more mirror. ObjectOstands 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 image distance i, and (f) the lateral magnification localid="1664204659535" m. (All distances are in centimeters.) It also refers to whether (g) the image is real localid="1664204665329" (R)or virtual localid="1664204670423" (V). (h) inverted localid="1664204676312" (I)or non-inverted localid="1664204682860" (NI)from O, and (i) on the same side of the mirror as the object Oor on the opposite side. Fill in the missing information, where, only a sign is missing, answer with the sign.

Short Answer

Expert verified

(a) The type of the mirror is concave.

(b) The focal length is +20cm.

(c) The radius of curvature is +40cm.

(d) The object distance is +10cm.

(e) The image distance is 20cm.

(f) The magnification ratio is +2.0.

(g) The image is virtual V.

(h) The image is non-inverted NI.

(i) The image is the opposite side of the mirror.

Step by step solution

01

The given data:

The mirror is concave.

The focal length of the mirror, f=+20cm

The object distance of the mirror, p=+10cm

02

The concept of the properties of a concave mirror:

A concave mirror is a diverging mirror with its reflective surface bugling opposite of the light source. Thus, as per the properties, the images formed by the concave mirrors can be both real and virtual based on the position of the object.

The focal length is positive if the mirror is a concave mirror. The focal length is negative if the mirror is a convex mirror. The image distance is positive if the image is a real image and is on the mirror side of the object.

Object distance refers to the separation between an object and the mirror's pole.

Magnification refers to the ratio of image length to object length measured in planes that are perpendicular to the optical axis.

Formulae:

The radius of curvature of a mirror,

r=2f ….. (i)

The mirror equation is,

1f=1i+1p ….. (ii)

The lateral magnification of an object,

m=hiho=ip ….. (iii)

Here, fis the focal length of the mirror, pis the object distance from the mirror, iis the image distance, hiis the height of the image, and hois the height of an object.

03

(a) Determining the type of mirror:

The mirror is a concave type as given in the problem. Hence, the type of mirror is concave.

04

(b) Determining the focal length:

As the given mirror is a concave mirror, the focal length value of the mirror will be positive in sign.

Thus, using the given data, the focal length of the mirror is given as:

f=+20cm

Thus, the focal length is +20cm.

05

(c) Determining the radius of curvature:

Thus, the radius of curvature of the mirror can be given using equation (i) as follows:

r=2f=(2×20cm)=+40cm

Therefore, the radius of curvature is +40cm.

06

(d) Determining the object distance:

The value of the object distance from the mirror is given by the given data in the table as follows:

p=+10cm

Hence, the object distance is +10cm.

07

(e) Determining the image distance i:

Now, the image distance can be calculated using the data in equation (ii) as follows:

120cm=1i+110cm

1i=120cm110cm=(1020)cm200cm2=10cm200cm2

i=-200cm210cm=20cm

Hence, the image distance is20cm.

08

(f) Determining the magnification ratio:

Thus, the lateral magnification of the mirror can be given using equation (iii) as follows:

m=ip=(20cm)10cm=+2.0

Therefore, the magnification ratio is +2.0.

09

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

From the calculations based in part (e), it is found that the image distance is negative in value. Thus, for an image distance to be negative, the image can be concluded to be virtual.

Hence, the image formed by the mirror is virtual V.

10

(h) Determining whether inverted or non-inverted:

The lateral magnification of the mirror is given to be positive value. Again, you know that the lateral magnification can be given as:

m=hiho=ip=+2.0

Thus, the image height needs to be positive that is possible only in an inverted image case.

Therefore, the image is non-inverted NI.

11

(i) Determining the position of the image:

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

Hence, the image is on the opposite side of the mirror.

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

An object is moved along the central axis of a spherical mirror while the lateral magnification m of it is measured. Figure 34-35 gives m versus object distance p for the rangepa=2cm and pb=8.0cm. What is m for p=14cm?

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?

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 , index of refraction n of the lens, radius localid="1662989860522" r1of the nearer lens surface, and radius localid="1662988669866" r2of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance iand (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real localid="1662988718474" Ror virtual localid="1662988727007" V, (d) inverted localid="1662988740117" Ifrom object or non-inverted localid="1662989876683" NI, and (e) on the same side of the lens as objectOor on the opposite side.

69 through 79 76, 78 75, 77 More lenses. Objectstands 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 or diverging , (b) the focal distance , (c) the object distance p, (d) the image distance , and (e) the lateral magnification . (All distances are in centimetres.) It also refers to whether (f) the image is real or virtual , (g) inverted or non-inverted from , and (h) on the same side of the lens asor 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.

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