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STEP 3   Since the image distance is negative, that is, q < 0, the image is virtual, located at a distance |q| behind the mirror. Therefore, the magnification M is M = −  q p  = −  −25.9 cm 1110 cm  = +0.023 

0.0234

 .(b) Since M > 0, the image is upright 

upright

 . Step 3 (c) In the previous step, we found the magnification M of the image. This magnification means that the image is  
Enter a number. times the size of the object.

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

An object is held 99.5 cm away from a mirror. The mirror hasafocal length of 72.9 cm.

a.) Determine the image distance.
1 cm

b.) Determine the magnification.
2

c.) If the object is 9.8 cm long, determine the image height.
3 cm

d.)Which of the following are true about the image? Choose allthatapply.

4

The image is virtual.

The image is upright.

The image is inverted.

The image is real.

a) The image distance

q =pf / p -f = 99.5*72.9 / 99.6-72.9 =272.69 cm

(b) Magnification m = -q/p = -272.69 / 99.5 = - 2.74

(c) magnification m = hf / hi

-2.74 = hf / 9.8

hf = -26.85 cm

(d) for lens image is real and inverted

In the diagram below the capacitor has a capacitance of79μF and the inductor has an inductance of 66 mH.Initiallythe capacitor has a voltage of 17 V. This circuit hasnoresistance.


A. How much charge does the capacitorinitiallyhave?
1 C

B. How much energy does the capacitorinitiallyhave?
2 J

C. What is the maximum current that willpassthrough the inductor?
3 A

Response Details:

a. Charge Q = C*V

= 79*10-6 * 17

= 1.343*10-3 C

b. Enegy storedinthecapacitor U = (1/2)*C * V2

= 0.5*79 * 10-6 * 172

= 1.14*10-2 J

c. Maximumenergyinindcutor = U

(1/2)*L*Imax2 = 1.14*10-2 0.5 * 66*10-3*Imax2 = 1.14*10-2

Imax = √(1.14*10-2 / 3.30 * 10-2)

= 0.5877 A

An object is held 99.5 cm away from a mirror. The mirror hasafocal length of 72.9 cm.

a.) Determine the image distance.
1 cm

b.) Determine the magnification.
2

c.) If the object is 9.8 cm long, determine the image height.
3 cm

d.)Which of the following are true about the image? Choose allthatapply.

4

The image is virtual. The image isupright. The image isinverted. The image isreal.



e.)Which of the following are true about the mirror? Choose allthatapply.

5

The mirror is concave. The mirror isconvex. The mirror isconverging. The mirror isdiverging.

Answer:

a) The image distance

q = pf / p -f =99.5*72.9 / 99.6-72.9 = 272.69cm

(b) Magnification m = -q/p = -272.69 / 99.5 =-2.74

(c) magnification m = hf /hi

-2.74 = hf /9.8

hf = -26.85cm

(d) for lens image is realandinverted

(e) ???

If a virtual image is formed 9.0 cm along the principal axis from aconvex mirror of focal length –15.0 cm, how far is the object fromthe mirror?

2. A woman looking in a makeup mirror sees her face at twice itsactual size and right-side up. If she is 26.0 cm from the mirror,what is its focal length?

3. Which of the following best describes the image of a concavemirror when the object is at a distance greater than twice thefocal point distance from the mirror?
a. virtual, upright and magnification greater than one
b. real, inverted and magnification less than one
c. virtual, upright and magnification less than one
d. real, inverted and magnification greater than one

4. A convex mirror with focal length of ?18 cm forms an image 12 cmbehind the surface. Where is the object located as measured fromthe surface?

5. A convex mirror with a focal length of ?18 cm forms an image 15cm behind the surface. If the object height is 1.2 cm what is theimage height?

6. An object placed 14 cm from a concave mirror produces a realimage 8.0 cm from the mirror. If the object is now moved to a newposition 18.0 cm from the mirror, where is the new image located asmeasured from the mirror?

7. When the reflection of an object is seen in a concave mirror theimage will:

8. A candle is 40.0 cm in front of a convex spherical mirror ofradius of curvature 70.0 cm. What are the image distance and themagnification, respectively?

9. An object is 16.0 cm from the surface of a spherical Christmastree ornament that is 8.00 cm in diameter. What is themagnification of the image?

10. A solid glass sphere with a radius of 6.00 cm and index ofrefraction of 1.50 has a small coin embedded 3.00 cm from the frontsurface of the sphere. For the viewer looking at the coin throughthe glass, at what distance from the front surface of the glassdoes the coin’s image appear to be located?

11. A glass block, for which n = 1.52, has a blemish located 4.5 cmfrom one surface. At what distance from that surface does the imageof the blemish appear to the outside observer?

12. An object of length 2.00 cm is inside a plastic block withindex of refraction 1.50. If the object is viewed from directlyabove, what is the length of its image?

13. An object is placed at a distance of 25 cm from a thin convexlens along its axis. The lens has a focal length of 10 cm. What arethe values, respectively, of the image distance andmagnification?
14. An object is placed at a distance of 50 cm from a thin lensalong the axis. If a real image forms at a distance of 35 cm fromthe lens, on the opposite side from the object, what is the focallength of the lens?

15. A contact lens is made of plastic with an index of refraction1.50. The lens has an outer radius of curvature of +2.0 cm and aninner radius of curvature of +2.5 cm. What is its focallength?

16. A Young’s double slit has a slit separation of 2.50 ? 10?5 m onwhich a monochromatic light beam is directed. The resultant brightfringes on a screen 2.00 m from the double slit are separated by2.30 ? 10?2 m. What is the wavelength of this beam? (1 nm = 10?9m)

17. A Young’s double-slit apparatus is set up so that a screen ispositioned 1.9 m from the double slits, and the spacing between thetwo slits is 0.040 mm. What is the distance between alternatingbright fringes on the screen if the light source has a wavelengthof 630 nm? (1 nm = 10?9 m)

18. The blue tint of a coated camera lens is largely caused by whateffects?

19. What is the minimum thickness of a glycerin film (n = 1.47) onwhich light of wavelength 650 nm shines that results inconstructive interference of the reflected light? Assume the filmis surrounded front and back by air.

20. Light of wavelength 560 nm is incident on a slit of width 0.150mm, and a diffraction pattern is produced on a screen that is 2.00m from the slit. What is the width of the central bright fringe? (1nm = 10?9 m)

21. Light of wavelength 610 nm is incident on a slit of width 0.25mm, and a diffraction pattern is produced on a screen that is 1.5 mfrom the slit. What is the distance of the second dark fringe fromthe center of the bright fringe? (1 nm = 10?9 m)

22. A multiple slit diffraction grating has a slit separation of2.50 ? 10?6 m. Find the wavelength of the monochromatic light thatwill have its second order bright fringe diffracted through anangle of 38.0°. (1 nm = 10?9 m)

23. A diffraction grating with 12 000 lines/cm will exhibit thefirst order maximum for light of wavelength 510 nm at what angle?(1 nm = 10?9 m)

24. A beam of unpolarized light in air strikes a flat piece ofglass at an angle of incidence of 58.0°. If the reflected beam iscompletely polarized, what is the index of refraction of theglass?

25. A beam of polarized light of intensity I0 passes through asheet of ideal polarizing material. The polarization axis of thebeam and the transmission axis of the sheet differ by 45°. What isthe intensity of the emerging light?

26. At what angle is the sun above the horizon if its light isfound to be completely polarized when it is reflected from the topsurface of a slab of glass (n = 1.65)?

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