Chapter 35
The Nature of Light and the Laws of Geometric Optics
EXAMPLES
Chapter 35: Nature of Light and Laws of
Geometric Optics; EXAMPLES
Example 35.1 Double Reflection
Two mirrors make an angle of 120o with each other. An
incident ray strikes the mirror M1 at an angle of 65o to
the normal. Find the direction of the ray after it is
reflected from mirror M2.
 The reflected ray is directed toward the
mirror M2. Making an angle of
90o  65o = 25o with the horizontal.
 From the triangle made by the first reflection and
the two mirrors: 180o  120o  25o = 35o (angle of
the first reflected ray with M2.
 Therefore, this ray makes an angle of 55o with the
normal to M2.
 From the law of reflection the second reflected ray
makes an angle of 55o with the normal to M2.
Example 35.2
Double Reflection (2nd part)
From Example 35.1 if the rays are extended behind the
mirrors, they cross at 60o, so that the overall change in
direction of the light ray is 120o (same angle between
mirrors). If the angle between mirrors is changed, is
the overall change in the direction of the light ray
always equal to the angle between the mirrors?
 NO! By using the law of reflection and the sum
of the interior angles of a triangle:
 = 180o  (90o  )   = 90o +    .
 From the highlighted triangle:
 + 2 + 2 (90o  ) =180o   = 2(  )
Example 35.2
Double Reflection, final
 The change in direction of the light ray is
the angle
 = 180o     = 180o  2(  ) 
 = 180o  2[  (90o +    )] 
 = 360o  2
 If  = 120o   = 120o
 If  = 90o   = 180o (RETROREFLECTION)
Applications:
Quick Quiz 35.2 Following the Reflected and
Refracted Rays
Ray  is the incident ray
Ray  is the reflected ray
Ray  is refracted into the Lucite block
Ray  is internally reflected in the block
Ray  is refracted as it enters the air from the block
Example 35.3
Angle of Refraction for Glass
Light of a wavelength 589 nm is refracted into a
crown glass slab with 1 = 30.0o.
(A). Find the angle of refraction 2 = ?
n1 = 1.00 and n2 = 1.52 (from Table 35.1)
From eqn. (35.8)
n2 sin 2  n1 sin 1
Then: 2 = sin-1(n1 / n2) sin 1 = 19.2o
The ray bends toward the normal, as expected
Example 35.3
Angle of Refraction for Glass, 2
(B). Find the speed of the light once enters the glass.
From eqn. (35.4)
(C). What is the wavelength of this light in the glass?
From eqn. (35.7)
Example 35.4
Measuring n from a Prism
The minimum angle of deviation (δmin) for a prism occurs when the
incident angle 1 is such that the refracted ray inside the prism makes the
same angle with the normal to the two prism faces.
Obtain an expression for the index of refraction of the prism material.
Example 35.4
Measuring n from a Prism, cont.
From the geometry:
2 

2
and 1   2     / 2   min / 2 
   min
2
From Snell’s Law (with n = 1 for air):
    min 
sin 1  n2 sin  2  sin 
  n sin   / 2  
2


    min 
sin 

2


n
 : Apex angle
sin   / 2 
Example 35.5
Find Critical Angle C
Find C for an Air-Water boundary.
n1 = 1.33 and n2 = 1.00
n2
1
sin  c 

 0.75 
n1 1.33
 c  sin
1
 0.75  48.8
o
Example 35.6
A view from the Fish’s Eye
A fish in a still pond looks upward toward the water’s
surface. What does it see?
Using the result from Example 35.5
 At an angle less than C (48.8°) the fish can see out
of the water
 At 48.8°the light has to skim along the water’s
surface before been refracted, so the fish can see the
whole shore of the pond.
 At an angle greater than C (48.8°) the fish sees a
reflection of the bottom of the pond.
Example 35.7
Light Passing Through a Slab
(A). A light beam passes from medium 1 to medium 2 with the latter medium being a thick slab
of material whose index of refraction is.
Show that the beam emerging into medium 1 from the other side is parallel to the incident beam.
Apply Snell’s law to the upper surface:
Apply Snell’s law to the lower surface:
Therefore:
sin  3 
n2
n1
 n1

 sin 1   sin 1
 n2

sin  2 
n1
sin 1
n2
sin  3 
n2
sin  2
n1
  3  1
Example 35.7
Light Passing Through a Slab, 2
(B). If the thickness t of the slab is doubled, what does happen to the offset distance d?
Find an expression for a from the yellow triangle: a  t / cos  2
Find an expression for d from the red triangle: d  a sin   a sin(    )
1
2
Combine these equations to find an expression for d in function of t:
Conclusion between d and t:
The angles (1 &  2 ) are only related to index of refraction
(constants), so d  t .
Therefore: If t doubles, so does d !
d
t
sin( 1   2 )
cos 2
Material for the Midterm
Material from the book to Study!!!
Objective Questions: 8-12
Conceptual Questions: 6-8
Problems: 6-8-15-16-20-37-39