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26 Nov 2019

An object is held 56.9 cm away from alens. The lens has a focallength of -69.8 cm.

a.) Determine the image distance.
1 cm

b.) Determine the magnification.
2

c.) If the object is 5.7 cm long,determine the image height.
3 cm

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

5
___The lens isdiverging.The lens is concave.___The lens isconvex.The lens is converging.

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Elin Hessel
Elin HesselLv2
22 Jul 2019

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Related textbook solutions

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

Related questions

An object is placed 20.0m from a converging lens with a focallength 15 cm. A concave mirror with focal length 10cm is located75cm to the right of the lens as shown in the figure. (Please referto: http://www.njaapt.org/Physics%20Competitions/Science%20League/Physics1Tests/April/Physics%201%20April%202006.pdf
pg. 5 # 23.
A) Determine the location of the final image.
B) If the height of the object is 1.0cm, what is the height ofthe image?
C) How would you describe the final image (size,orientation)?

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) ???

b) In procedure 1 (part 4), light shines on a convex lens. Some ofthe light refracts through the lens, forming a real focus F1. Someof the light reflects off the front surface (which acts like aconvex mirror), and forming a virtual focus F2.

- If the index of refraction of the lens material increases, F1will
increase
remain the same
decrease

- If the index of refraction of the lens material increases, F2will
decrease
remain the same
increase

c) In procedure 1: the laser box is placed 10.5 cm from the convexlens, and you find the focal length of the lens is 8.57 cm. If thedistance from the lens to the laser box is now 20.1 cm, , the focallength of the lens will now be
cm

d) Suppose you do procedure 1 and determine that the magnitude ofthe focal length of the concave lens is f. If the measurement isnow done underwater (n = 1.33) instead of in an air atmosphere, themagnitude of the focal length will
decrease
increase
remain the same

2. –/3 points Notes
In procedure 2, suppose the convex lens is placed at the 35.4 cmmark, and an image forms on the screen at the 63.8 cm mark.Find:

- the focal length of this lens. cm
- the absolute value of the magnification of this lens
- the height of the image cm

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