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GGR205H1 Midterm: Midterm review

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redalligator469
28 Oct 2011
School
UTSG
Department
Geography
Course
GGR205H1
Professor
Alison Smith

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Document Summary

The unconsolidated mineral or organic material on the immediate surface of the earth that serves as a natural medium for the growth of plants. Solum: the upper and most weathered part of the soil profile, the a,e and b horizons. Regolith: the unconsolidated mantle of weathered rock and soil material on the earth surface. Medium for plant growth, recycling system for nutrients and organic wastes, modifier of the atmosphere, habitat for soil organisms, engineering medium, system for water supply and purification, Soil scientists often dig a large hole called soil pit, usually several meters deep and about a meter wide to expose soil horizons for study. The vertical section exposing a set of horizons in the wall of a pit: what are the five soil horizons most commonly found in humid forested regions of. Physical chemical difference: define an essential element.

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Related Questions

You will be utiizing a basicoptic system.

The index of refraction of theplastic cylindrical lens depends on wavelength. Choose an anglewhich gives a good color separation and measure the index ofrefraction for blue and red light.

Question:How big is the difference in indices of refraction? How good is theapproximation that n is constant for thismaterial?

If this question doesnt make sense..then just answer ittheoretically speaking. That is how would we go about checkingwhether or not it conforms with the n as the constant for thematerial... Thanks!

Background info:

ThePhysics

The Law ofReflection

The law of reflection states that for a plane mirror, the angle ofincidence is equal to the angle of reflection, and that the tworays lie in a plane.

r

normal

. A light ray perpendicular to the surface has an angle ofincidence 0°.

The Law of Refraction –Snell’s Law

A ray of light passing from one substance (index ofrefraction

n1)into another (index of refraction n

2) across a plane boundary will be refracted. The angle the raymakes in one medium will be related to the angle in the secondmedium by the equation below, and that all rays lie in aplane.

(2)

Total InternalReflection

A ray incident on a medium with lower index of refraction willbe

totally internally reflected if sin θ1> n2/n1.There will be no transmitted beam for this incident angle. Ameasurement of the critical angle for total internalreflection can be used to determine the index of refraction of the substance.If n

2 = 1 (air), then the critical angle is:

sin θc = 1/

n

1. (3)

Polarization byReflection – Brewster’s Angle

If the angle between the reflected and refracted rays is 90°,then

tan θ

B = n2/n

1. (4)

At this angle, only light with polarization out of the plane ofreflection and refraction will be reflected. The reflected lightwill be completely polarized.

Figure3.

Refractionof Light Rays

Figure2.

Incidentand Reflected Rays

(1)

Always measure the angles with respect to the surface

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)?

Materials fracture when the force per unit area, called STRESS,exceeds a critical value. In general, the critical stress dependson the direction as well as the magnitude of the force. Forexample, bones break differently when subjected to torsional(twisting) stress than when subjected to compressional (orsqueezing) stress. Let us consider compressional stress. Wet humanbone for 20-39 year olds has an ultimate compressive strength of ~16x104Ncm2 [8]. Consider a person who falls from a height to theground. Upon landing, the person's kinetic energy is converted towork. If the person decelerates over a distance h, we can estimatethe average force during the collision by

If a person lands stiff-legged on a hard surface and doesn'tbounce, the deceleration occurs over a very short distance, h ~ 1cm. Since the force is transmitted up the leg, the stress isgreatest where the cross-sectional area is least, i.e., the tibiajust above the ankle [8].
The following lists some different conditions that may occur duringlanding and physical explanations for why they make bone fracturefrom a fall more or less likely [8]:

(a) bounce [more likely. The impulse is twice as large for anelastic collision as for an inelastic collision.]
(b) bent knees [less likely. Bending the knees increases the timeand distance over which deceleration occurs, which reduces theforce.]
(c) landing in loose earth or sand [less likely. This increases hwhich reduces F.]
(d) turning the body to land on the side [less likely. Thisincreases h. It also distributes the force widely over the body(increases A), thereby reducing the stress.]

So, why can a properly executed karate kick break a concrete blockwithout fracturing bones [16]? First, bone is a very strongmaterial. Its ultimate compressive strength is approximately 40times larger than concrete. Second, contact is made with the edgeof the foot. This concentrates the force into a small area of thetarget and reduces the likelihood of bending a bone to the point offracture. Third, the collision with the target is essentiallyinelastic and extends over several milliseconds, so the peak force,though large, does not exceed the strength of the bone. If A(cross-sectional area of the tibia) ~ 3.3 cm2, compute how far a 69kg person can fall and land stiff-legged on both legs withoutbreaking a bone. Assume F is split evenly between two legs. (Hint:F<or =2(theta)A )