Physical Foundations in Engineering II Spring 2003
Due: Monday, January 27, 2003, at 9 am
Wolfson and Pasachoff, Ch. 35, Question 6 (p.932)
First, let’s assume that the index of refraction for all frequencies of visible light is constant, i.e. n red
= n orange
= n yellow
= … = n violet
. Then, the only dispersive medium (i.e. the one in which different wavelengths of light have slightly different indices of refraction) in the problem is that of glass slab with parallel surfaces. In this case, the light entering the slab and that exiting it must be parallel to one another, which is a direct consequence of Snell’s Law.
Since inside the glass slab the white light is dispersed, light passing through it experiences lateral displacement , as shown in the figure below, and stays dispersed on exiting the slab.
If the lateral displacement of different wavelengths of light upon exiting the slab were small enough to be visually resolved, one would observe “a mixture of colors”, i.e. white light exiting the slab. However, if the lateral displacement of the exiting light is large enough to be visually resolved, one could actually observe dispersion of light. The value of lateral displacement is defined both by the width of the glass slab and by the angle of the incidence of the original light, as follows: d
w
 incident
n i
n
(i-1)
where d is the lateral displacement between the neighboring wavelengths of light
 i
and

(i-1)
, w is the width of the glass slab, and n i
is the index of refraction of the i th wavelength of light.
Figure taken from http://www.physics.otago.ac.nz/Physics100/simulations/Gamelan/java/dispslab/index.html
Wolfson and Pasachoff, Ch. 35, Question 7 (p.932)
In Figure 35-15(a) a glass prism redirects light through total internal reflection. Given that the angle of incidence on the second interface between prism and air is 90° and the angle of reflection from this interface is 45°. For total internal reflection to take place with the critical angle of 45°, the following condition must be true:
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Physical Foundations in Engineering II Spring 2003 sin
  c n air n glass
, but we have sin 45

0.707
 n air n glass

1
1.25

0.8
.
Therefore, the light entering prism of 1.25 index of refraction cannot experience total internal reflection, as shown in the figure.
Wolfson and Pasachoff, Ch. 36, Question 1 (p.965)
Light rays diverge from the position of a virtual image just as they do from an actual object.
Thus, a virtual image can be as easily seen as any object can. Of course, your eye would have to be placed near the axis of the mirror in order to intercept the light rays.
4.
Wolfson and Pasachoff, Ch. 36, Problem 2 (p.965)
When you face a mirror to take a photograph with an autofocus camera that sets its focus by bouncing ultrasound waves off the objects at which it is pointed, it will be focused on the surface of the mirror. As the goal is to take a picture of your own image, the distance between the camera and your image is twice of that between your camera and the mirror. As a result, the only way to obtain “sharp” photograph of your image would be to stand very close to the mirror, such the difference between the distances between you and the mirror and that between you and your image is not very large.
Wolfson and Pasachoff, Ch. 36, Question 4 (p.965)
The only time when an object and its image in the concave mirror will be of the same size is when the object is located at the center of curvature. This is a direct result of the mirror equation and the definition of the magnification, i.e.:
M l l 2 l
Rl

R l
1

1 .
The only way for the above expression to be true is for the object to be located distance R away from the mirror, i.e. at the center of the curvature.
Please, see WebAssign for Solutions.
Due: Tuesday, January 28, 2003, at 9 am
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Physical Foundations in Engineering II Spring 2003
Due: Tuesday, January 28, 2003, in class
Wolfson and Pasachoff, Ch. 35, Problem 25 (p.935)
Light propagating in a medium with refractive index n
1
encounters a parallel-sided slab with index n
2
. On the other side is a third medium with index n
3
 n
1
.
Show that the condition for avoiding internal reflection at both interfaces is that the incidence angle at the n
1
-n
2
interface be less than the critical angle for an n
1
-n
3
interface. In other words, the index of the intermediate material doesn’t matter.
Since medium-2 has parallel interfaces with media-1 and 3, the angle of refraction at the 1-2 interface equals the angle of incidence at the 2-3 interface, as shown. (The normals to the interfaces are also parallel, so the alternate angles, marked

2
, are equal.) Thus Snell’s law implies n
1 sin

1
 n
2 sin

2
 n
3 sin

3
, so that the angles in media-1 and 3 are related as if media-2 were not present. Of course, there are conditions on the intensity of the light transmitted through medium-2 which do depend on n
2
and the critical angle, if any, at the 1-2 interface. If n
2
 n
1
, the phenomenon of frustrated total reflection (i.e., transmission of light for angles greater than the critical angle) may occur if the thickness of the slab is on the order of a few wavelengths of the incident light.
Wolfson and Pasachoff, Ch. 35, Problem 33 (p.935)
A scuba diver sets off a camera flash a distance h below the surface of water with refractive index n. Show that light emerges from the water surface through a circle of diameter 2 / n
2 
1.
Light from the flash will strike the water surface at the critical angle for a distance r
 h tan
 c
from a point directly over the flash. Therefore, the diameter of the circle through which light emerges is
2 r

2 h tan
 c
.
But sin
 c

1/ n (Equation 33-5 at the water-air interface), and tan
2  c

(csc
  c

1 )

1
(a trigonometric identity), so we can write 2 r

2 / n
2 
1.
Wolfson and Pasachoff, Ch. 35, Problem 57 (p.937)
Show that a three-dimensional corner reflector (three mirrors in three mutually perpendicular planes, or a
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Physical Foundations in Engineering II Spring 2003 solid cube in which total internal reflection occurs) turns an incident light ray through 180°, so it returns
in the direction from which it came. Hint: Let q
 q x
î
 q y j
  q z
be a vector in the direction of propagation. How does this vector get changed on reflection by a mirror in a plane defined by two of the coordinate axes?
A single plane mirror reverses the direction of just the normal component of a ray striking its surface.
For example, a ray incident in the direction
 
î cos
 x
 j
 cos
 y
 cos
 z
, on a mirror normal to the x -axis, is reflected into the direction

î cos
 x
 j
 cos
 y
 cos
 z
.
(In our notation,

is a unit vector, and cos
2  x
 cos
2  y
 cos
2  z

1 .
) If the ray also strikes mirrors which are normal to the y - and z -axes, as in a corner reflector, it emerges in the direction
    
, or opposite to the initial direction. In order to strike all three mirrors, the direction cosines of the incident ray must have magnitudes greater than some minimum non-zero value, depending on the size of the reflector (i.e., cos
 i

0 for i
 x y and z ).
Wolfson and Pasachoff, Ch. 35, Problem 61 (p.937)
Fermat’s principle states that the path of a light ray between two points is such that the time to traverse that path is an extremum (either a minimum or maximum) when compared with the times for nearby paths.
Consider two points A and B on the same side of a reflecting surface, and show that a light ray traveling from A to B via a point on the reflecting surface will take the least time if its path obeys the law of reflection. Thus, the law of reflection (Equation 35-1) follows from Fermat’s principle.
Take the x-z plane to be the reflecting surface, and the x-y plane to contain the points A and B , with the y -axis through A . Suppose a ray traveling from A to B via the surface is reflected at point P x 0 z , as shown. (It is assumed that the path of a ray between two points in the same medium, with no intervening reflections or refractions, is a straight line, which also follows from Fermat’s principle.)
Since the ray propagates in only one medium, the time to traverse the path APB is proportional to the total distance ( t

D =v ), where D

AP

PB
 x
2  y
2
A
 conditions for D to be an extremum (a minimum in this case) are z
2 
  z
( x

B
 x )
2
0 and
 y
2
B
 
 x z

2
.
The
0. The first condition requires that ( z = x
2  y
2
A
 z
2
)

( z = same plane, called the plane of incidence.) With
( x
B
 x )
2  y
2
B
 z
2
)

0 , which is satisfied only for z

0.
(Thus, the incident ray, the normal to the surface, and the reflected ray all lie in the z

0, the second condition requires that
( x = x
2  y
2
A
)

( x
B
 x ) = ( x
B
 x )
2  y
2
B

0 .
But sin
  x = x
2  y
2
A
, and sin
  
( x
B
 x ) = ( x
B
 x )
2  y
2
B
, therefore sin
  sin
  
0 which is the law of reflection. (Note: If A and
B are on opposite sides of the x-z plane in media where the velocity of light is different, differentiation of t

( AP =v )

( PB =v

) leads to (sin

=v )

(sin

=v
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Physical Foundations in Engineering II
A (0, y
B (0, y
Spring 2003
P ( x ,0, z )
a) Wolfson and Pasachoff, Ch. 35, Problem 36 (p.935)
White light propagating in air is incident at 45

on the equilateral prism of Fig. 35-38. Find the angular dispersion

of the outgoing beam, if the prism has refractive indices n red

1 582 , n violet

1 633 .
This geometry for refraction through a prism is treated in Problem 55. The difference in the directions of the exiting red and violet light is the dispersion. For the given values of
  and
60
 r

,


1
45 7 dispersion is



45
 
 v
, n
 r


, while r
1 582

2 ,
(or v

,

1 ,
and v
  n
25 6

, v
1 , r


2
1 633
, v

, we find:
34 3

,

1 , v


2 , r

67 1

26 5

,
and


2 v
, r


33 5

,

1 , r

60 7

,
)

6

(Note: We did not need to calculate the deflection, but did so because this is the customary parameter in prism optics.) b) Wolfson and Pasachoff, Ch. 36, Problem 2 (p. 966)
Two plane mirrors occupy the first four meters of the positive x- and y-axes, as shown in Fig. 36-44. Find the locations of all images of an object at x

2 m, y

1 m.
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Physical Foundations in Engineering II
Solution
Spring 2003
I I I
I I I I I I
In addition to the two direct images in each mirror (one reflection) at I
I x

( 2 m ,

1 m ), a multiple image (two reflections) also appears at I xy y is the image in one mirror of the direct image in the other.) Ray tracing confirms this, as shown on figure above. No more than two reflections are possible for perpendicular mirror planes, so these are all the images.
 
2 1 ) and
2 m ,

1 m).
(The latter
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