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Textbook Guide Physics: Mount Everest, Solar Cell, Sodium-Vapor Lamp

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grayraven524
1 Dec 2016
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Physics
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PHY131H1
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PHY131H1 Full Course Notes
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PHY131H1 Full Course Notes
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Document Summary

The wavelength of light is thus expressed as represents the wavelength in a vacuum. The phase difference between two waves can change if they pass through different materials with different indexes of refraction. The light leaving these slits flares out or diffracts, and interference occurs in the region beyond the screen. A fringe pattern forms on a viewing screen. This behavior was visible in other types of waves, thus proving light was a wave: the path length difference determines the appearance of each point in the viewing screen when a young"s double-slit interference experiment is performed. The path length difference is expressed as: the conditions for maximum and minimum intensity are expressed by: represents the angle the light path makes with a central axis. represents the slit separation. 35-3 interference and double-slit intensity: coherence in light (and therefore perceptible interference) is extant when the phase difference remains constant when two light waves meet at a point.

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Related textbook solutions

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

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Related Questions

Huygens' principle (first described by the Dutch scientist Christiaan Huygens in 1678) uses geometry to determine the shape of a wavefront at a time t, given some initial wavefront at an earlier time. This can be constructed by imagining that the initial wavefront is a source of wavelets that propagate from each point at the speed of light. From this principle, one can determine the angles of reflection and refraction by considering the speed of light at each point on the wavefront. In this problem, we will explore some of these concepts.

Part A

First, let's look at a plane wave that is incident on a flat piece of material with an index of refraction n. Part of the light is transmitted through the material and part of it is reflected. For the moment, let's just look at the part that is transmitted. At time t=0, the wavefront is a distance d away from the surface of the material. At time t=t1, the wavefront is at the position of the material interface. Use the Huygens' principle to determine how far into the material (d') the wavefront has propagated by time t=2t1.

Express your answer in terms of the variables n, t1, and the speed of light in a vacuum c.

Note that there would be no change in the direction of the wavefront at any point because the wavefront encountered the material interface at the same time at all points. Thus, all of the individual wavefronts all propagated at the same speed, thereby maintaining the flat wavefront.

Now, instead of having a flat wavefront propagating normal to the material interface, we have a flat wavefront propagating toward the material at an angle of 55° to the axis perpendicular to the material interface. In this part, we will look at the relative positions of a few points—A, B, and C—on the wavefront to illustrate Huygens' principle. (Intro 1 figure) Point C touches the vacuum/material interface at time t=0 whereas point B is a distance d and point A is a distance 2d away from the interface.

Part B

What is the time tB it will take for point B of the wavefront to encounter the vacuum/material interface?

Express your answer numerically to two decimal places in units of t1 (the time it takes light to travel a distance d in a vacuum).

It should be noted that the accuracy of the method increases the smaller the distance is before making a new wavefront and redoing the wavelets. If you tried to just draw a large wavelet from point B, then it would hit the surface after only traveling a distance d. This large wavelet would make it difficult to visualize how to add up all of the other wavelets to make a coherent wavefront. As a result, one should just draw a line perpendicular to the wavefront at point B until it hits the surface, at which point the wavelets will change owing to the new index of refraction.

 

Part C

How far did point C go into the material interface in the time t=tB that it took for point B to get to the interface? For this part we are looking for the distance traversed by the point C in the material (dC).

Give your answer in terms of the time tB, c, and n.

 

Part D

What is the new angle θ at which point C of the wavefront is propagating (relative to a line perpendicular to the vacuum/material interface)? Try to use the fact that you have a spherical wavefront propagating from t=0 at the point where C met the vacuum/material interface until time tB when the wavefront at point B reached the interface.

Express your answer in terms of inverse trigonometric functions and n.

 

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)?