Ch 14 Refraction

Updated 5/26/07

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14.1 Refraction, pg 488

Practice A, Snell’s Law, pg 493
[1] 18.5^o     [2a] 27.5^o   [b] glycerin (n = 1.47)   [c] 12.5^o    
[3] 1.47    

Section 14.1 Review Q, pg 493: 
1] 16.7^o 
2a] away     [b] toward     [c] toward     [d] away
3] 6.14^o
4] a & d

Ch 14 Review Q, pg 514: 
1] No, not when ni > nr, or when wave hit boundary at 90^o.
2] Its wavelength gets shorter, velocity slows, frequency is constant
3] since n = c/v--->v = c/n, i.e., the velocity in a medium is inversely proportional to the index of refraction.
4] Light rays from the bottom bend away from the normal so that the image is closer to the observer.
5] For refraction to occur:  ∢ ≠ 0^o or 90^o, ni ≠ nr, both media must be transparent.
6] X, since it bends more it has a higher n and moves slower.
7] The image of the or underwater is closer to the observer.
8] Behind, since the image bends away from the normal you will aim too far.  Draw the diagram and see.
9] Since helium’s n = 1.03 it bends light very little & resembles air
10] 30.3^o  
11] 26^o  
12] 25.5^o 
13] 30.0^o, 19.5^o, 19.5^o, 30.0^o 
14] θ1 = 30.4^o, θ2 = 22.3^o

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14.2 Thin Lenses, pg 494

p = Do, q = Di, h = Ho, h’ = Hi

Practice B, Lenses, pg 501
[1] 20.0 cm, M = -1.00; real inverted image
[2] -30.0 cm, M = 3.00; virtual, upright image
[3] -6.67 cm, M = 0.333; virtual, upright image
[4a] Do = 2.0 cm, M = 1/5
  b] Do = 5.0 cm, M = -1.4
  c] Di = -2.4 cm, M = 0.60
  d] f = -5.0 cm, Di = -2.5 cm

Section 14.2 Review Q, pg 505: 
1] real, inverted
2a] virtual   [b] real   [c] virtual
3] 9.3 cm
4] -1.3
5] Di = -3.3 cm, Hi = 0.33 cm
6] The length of the telescope is slightly shorter than fo + fe

Ch 14 Review Q, pg 514: 
15] Converging
16] Rays are refracted parallel to one another.
17] Object:  [a] outside F   [b] inside F   [c] outside F
             [d] inside F    [e] inside 2F  [f] outside 2F
18] [a] Never    [b] Always  [c] Never     
    [d] Always   [e] Never   [f] Always
19] Light from a point image at infinity will enter the lens parallel
    to the principal axis and converge at F;  Focus sunlight on the
    ground and measure F
20] Longer
21] converging lens; outside
22] The image produced by the objective lens is inside Fe. 
    Therefore, the eyepiece produces a virtual image of a real,
    inverted image.
23] Yes, because n_ice > n_air
24] [a] -13.3 cm, M = 0.332; virtual, upright
    [b] -10.0 cm, M = 0.500; virtual, upright
    [c] -6.67 cm, M = 0.667; virtual, upright
25] 3.40; upright
26] [a] 40.0 cm, M = 1.00; real, inverted
    [b] -20.0 cm M = 2.00; virtual upright

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14.3 Optical Phenomena, pg 506

Practice C, Critical Angle, pg 508
[1] 42.8^o
[2] 64.82^o
[3] 49.8^o
[4] Diamond (24.4^o); cubic zirconia (27.0^o)

Section 14.3 Review Q, pg 511: 
1] 79.11^o 
2] b, d, e
3] green
4] evening, because the sun must be behind him to see the rainbow. 

Ch 14 Review Q, pg 515: 
27] No because n_air < n_water 
28] The air next to the ground must be hotter than the air above it
29] Light from the blue sky is refracted upward as it passes close to
    the ground.
30] Violet light is deviated more than red light as it passes
    through the drops of water.
31] [a] chromatic   [b] spherical   [c] spherical   [d] spherical  
32] As the density changes, the speed of light through it changes. 
    Thus, the light is continually refracted as n changes.
33] Rays initially moving upward are bent because temperature
    increases with height.
34] The angle of reflection does not depend on the wavelength of the
    light.
35] Light entering the diamond is dispersed.  Each color is totally
    internally reflected until the incident angle is less than the
     critical angle.
36] 42.8^o 
37] [a] 31.3^o   [b] 44.2^o   [c] 49.8^o  
38] It will be totally internally reflected because
    ∢i (45^o) > ∢c (41.1^o)

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Mixed Review, pg 516

     39] 1.31       [40] 51.9^o      [41] 1.62; Carbon Disulfide
42a] 6.00 cm   [b] 53.4^o       [43] 7.50 cm     [44] 8.55 cm
45a] 6.00 cm  
  b] A diverging lens cannot form an image larger than the object.
46] -80.0 cm                   [47a]  3.01 cm   [b] 2.05 cm    
48]  Do = 120 cm; M = 0.250    [49] Blue:  47.8^o, red:  48.2^o 
50] 16.5^o      [51]48.8^o       [52] 67^o         [53] 4.54 m
54] 110.6^o     [55] 10F/9      [56a] 40.8o      [b] 60.6^o
57a] 24.7o     [b] It will pass through the bottom surface because
                   ∢i < ∢c (∢c = 41.8)
58] 1.3        [59] 1.38       [60] 53.4^o  

Standardized Test Prep, pg 520

     1] D  [2] H  [3] A  [4] J  [5] B  [6] J  [7] A  [8] F [9] C  [10] H 11] The focal point of the objective must lie within the focal point
    of the eyepiece 
12] 27.0^o   [13] 50.0^o   [14] Skip   [15] 15 cm   16] 1.5  
17] 4.2 cm  [18] Virtual & Upright

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14.1 Refraction Notes

Objective:  Recognize situations in which refraction will occur.

1    What are the conditions that must be met for refraction to occur?
a] Both mediums must be transparent.
b] The mediums must have different indices of refraction.
c] The wave must enter a new medium at an angle greater than the critical angle and less than 90^o. 

2    State the 3 minor refraction laws.
a] When a wave moves into a new medium at 90^o to its surface no bending occurs.
b] When a wave moves into a new medium at an angle below its critical angle it is completely reflected and no bending occurs.
c] The amount of bending (refraction) increases with frequency.

Objective:  Identify which direction light will bend when it passes from one medium to another.

3    State & diagram the 2 main refraction laws.
a] When a wave moves into a more dense medium (from low to high index of refraction [n]) it moves toward from the normal.
b] When a wave moves into a less dense medium (from high to low index of refraction [n]) it moves away from the normal.

Practice A, pg 493

Objective:  Solve problems using Snell’s Law

1    Given:  ∢i = 25.0^o
Air (n = 1.000) to water (n = 1.333, see Table 1 pg 490): 
   ray bends toward normal----> ∢r < 25.0^o  
n1 Sin(i) = n2 Sin(r)
1 Sin(25) = 1.333 Sin(r)--->Sin(r) = 1 Sin(25)/1.333 = 0.3170
r = Sin^-1(0.3170) = 18.5^o 

2    a]  Given:  ∢i = 25.0^o
Flint Glass (n = 1.66) to Crown Glass (n = 1.52, see Table 1 pg 490):
    Diagram:  ray bends away from normal----> Estimate ∢r > 25.0^o 

n1 Sin(i) = n2 Sin(r)
1.66 Sin(25) = 1.52 Sin(r)--->Sin(r) = 1.66 Sin(25)/1.52 = 0.4615
r = Sin^-1(0.4615) = 27.5^o 

b]  Given:  ∢i = 14.5^o , ∢r = 9.8^o
            Air (n = 1.000) to ? (see Table 1 pg 490): 
            Diagram: ray bends toward normal---->Estimate n > 1.00

n1 Sin(i) = n2 Sin(r)
1.000 Sin(14.5) = n2 Sin(9.8)--->n2 = 1.00Sin(14.5)/Sin(9.8) = 1.471
 
n2 = 1.471--->See Table 1, pg 490--->Medium 2 = Glycerine

c] Given:  ∢i = 31.6^o
Air (n = 1.000) to diamond (n = 2.419, see Table 1 pg 490): 
   Diagram:  ray bends toward normal----> Estimate ∢r < 31.6^o 

n1 Sin(i) = n2 Sin(r)
1 Sin(31.6) = 2.419 Sin(r)--->Sin(r) = 1 Sin(31.6)/2.419 = 0.2166
r = Sin^-1(0.2166) = 12.5^o 

3    Given:  ∢i = 40.0^o ∢r = 26.0^o
         Air (n = 1.000, see Table 1 pg 490): 
         Diagram:  ray bends toward normal--->Estimate n2> 1.000
Find n2 = ?

n1 Sin(i) = n2 Sin(r)
1 Sin(40.0) = n2 Sin(26.0)---> n2 = 1.00Sin(40.0)/Sin(26.0) = 1.466
r = Sin^-1(0.3170) = 18.5^o 

14.1 Ch Review:  pg 514

10, 11, 12 Ask for the angle of refraction.  See Practice B Q 1,2,3

13   Since there are two surfaces there are 2 ∢i and 2 ∢r.
Since the glass is flat ∢r1 = ∢i2 and ∢r2 = ∢i1.

14   Given:  ∢i = ?, Air--->n = 1.000, Linseed Oil--->n = 1.48, ∢r = 20^o 
n1 Sin(i) = n2 Sin(r)
1 Sin(i) = 1.48 Sin(20^o)
Sin(i) = 0.5062---> i = sin^-1(0.5062) = 30.4^o  

Given: ∢i = 20, Linseed Oil--->n = 1.48, Water---> n = 1.333, ∢r = ?
n1 Sin(i) = n2 Sin(r)
1.48 Sin(20) = 1.33 Sin(r)---> Sin(r) = 1.48 Sin(20)/1.33 = 0.3797
r = sin^-1(0.3797) = 22.3^o  

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14.2 Thin Lenses Notes

Objective:  Use ray diagrams to find the position of an image produced by a converging or diverging lens, and identify the image as real or virtual

Drawing Instructions (Diagram #2 to #6, Table 3, pg 497)
a] Red = Object & Incident Rays, Blue = Reflected Rays & Image
b] Object always 3 rows high
c] Midline always at 3”
d] Draw a red parallel incident ray from the Object tip to midline
e] Draw a blue refracted ray from the midline through the right focal
   point.
f] Draw red incident ray in line with the left focal point to midline
g] Draw a blue parallel refracted ray from the midline to page edge.
h] Draw image from axis to the intersection of the 2 refracted rays.
i] Label & Measure:  Do, Ho, Di, Hi, & F
j] Find the image size & type.

Practice B, pg 501

Objective:  a] Solve problems using the thin-lens equation
       b] Calculate the magnification of lenses.

1    Converging Lens, Given:  Focal Length = 10.0 cm, Do = 20.0 cm = 2F
Find:  Di, Magnification & Image Description

1 = 1  + 1   --->1 - 1  = 1      
F   Do   Di      F   Do   Di

1  - 1  = 1  ----> 2  - 1  = 1 ----> 1  = 1 --->Di = 20 cm
10   20   Di       20   20   Di      20   Di

Magnification = Hi But Hi = Di = 20 = 1                           
                Ho     Ho   Do   20

Diagram (see #2 pg 497):  Object 3 rows high, Scale 1 inch = 10 cm
Place midline at 3”, draw lens 4 rows up & down from the midline.
   Image is real, inverted, same size as object.

2    Given:  Converging/Convex Lens; Focal Length = 15 cm, Do = 10 cm
Find:  Di, Magnification & Describe the Image
Since Do < F the diagram looks like Fig 6, Table 3, Pg 497
The image is erect, virtual and enlarged.

1 = 1  + 1   ---->1  = 1  + 1  ----> 1  - 1  = 1
F   Do   Di       15   10   Di       15   10   Di

----> 2  - 3  = 1 ----> -1  = 1 --->Di = -30 cm (The - sign indicates
      30   30   Di       30   Di                 a virtual image.)

Magnification = Hi But Hi = Di = 30 = 3                           
                Ho     Ho   Do   10

Diagram (see #6 pg 497):  Object 3 rows high, Scale 1 inch = 10 cm
Place midline at 3”, draw lens 10 rows up & down from the midline.

3    Given:  Diverging/Concave Lens, Do = 20 cm, F = -10 cm
Find:  Di, Magnification & Describe the Image
Since it is a diverging lens the image is erect, virtual, & reduced.
1 = 1  + 1   ---->-1  = 1  + 1  ----> -1  - 1  = 1
F   Do   Di        10   20   Di        10   20   Di

---> -2  - 1  = 1 ----> -3  = 1 --->Di = -20 cm (The - sign indicates
      20   20   Di       20   Di           3     a virtual image.)

Magnification = Hi But Hi = |Di| =  20/3 = 1/3                    
                Ho     Ho   |Do|     20

Diagram (see Fig 7, pg 498): Object 3 rows high, Scale 1 inch = 10 cm
Place midline at 3”, draw lens 6 rows up & down from the midline.