8. How long would a closed organ pipe need to be to produce a frequency of 400 H

Question

8. How long would a closed organ pipe need to be to produce a frequency of 400 Hz?

9. A concave mirror has a focal length of 30 cm. What kind of image is produced by an object located 60 cm from the mirror? Where is it located and what is its magnification?

10. A 15-cm ruler is palced 30 cm in front of a concave mirror with a focal length of 20 c,. Where is the image located and how big is it?

11. A 15-cm ruler is palced 75 cm from a converging lens with a focal length of 60 cm. Where is the image located and how big is it?

12. A tourist in Yellowstone National Park takes a photograph of a buffalo that is 40 m away and 2.2 m tall. If the lens has a focal length of 50 mm, how large is the image of the buffalo on the film?

13. Upon entering material X, light slows down by 30 percent compared to its speed in vacuum. What is the index of refraction of material X?

14. What is the thinnest soap film that will strongly reflect yellow light with a wave length of 589 nm?

Please show work and explain.

Explanation / Answer

Number 8)

v = f(wavelength)

343 = 400(wavelength)

wavelength = .8575 m

Pipe length is 1/4th of that

.8575/4 = .214 m

Number 9)

1/f = 1/p + 1/q

1/30 = 1/60 + 1/q

q = 60 cm in front of the mirror

The image is real, inverted and its magnification is 1 since image and object distances are the same

Number 10)

1/20 = 1/30 + 1/q

q = 60 cm

The image is 60 cm in front of the mirror

The height...h'/h = -q/p

h'/15 = -60/30

h' = -30 cm (You can ignore the negative, its sign convention meaning inverted)

Number 11)

1/60 =1/75 + 1/q

q = 300 cm behind the lens

For the height

h'/15 = -300/75

h' = -60 cm (You can ignore the negative, its sign convention meaning inverted)

Number 12)

1/5 = 1/4000 + 1/q

q = 5.01 cm

h'/h = -q/p

h'/2.2 = -5.01/4000

h' = -.00275 m which is -2.75 mm -- Again, ingorne the negative for sign convention if needed.

Number 13)

n = c/v

n = 1/.7

n = 1.43

Number 14)

2nt = (m + .5)(wavlength)

2(1.33)(t) = (.5)(589 X 10-9)

t = 1.11 X 10-7 m which is 111 nm

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