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ECE181 Homework Set1 Solution
April 9, 2015
Problem 1:Mirrors
There are two ways to solve this problem. The first idea of this problem is
that light incident from toe can be received from the eye after reflected by
planar mirror.
The light path is shown in fig1. We know that ✓1 = ✓2 , so h1 = h2 = h0 . The
height of the mirror is:
ho =
H
= 1m
2
(1)
and this answer doesn‘t depend on the distance of the mirror.
Figure 1: Problem 1a
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The second idea is that the person needs to see its image from the mirror
and the boundary of the mirror limits its vision, as shown in fig2. We can
also get that h0 = 1m.
Figure 2: Problem 1b
Problem 2: Fermat‘s Principle and Snell‘s law
To use Fermat‘s principle we want to express the optical path (OP) in terms
of a single independent variable x, and set dOP
= 0.
dx
Define x , height of intercept point B, as shown in fig??
q
l1 = d21 + x2
q
l2 = d22 + (d
x)2
(2)
q
q
2
2
OP = n1 l1 + n2 l2 = n1 d1 + x + n2 d22 + (d
q
q
dOP
d
d
2
2
=
n1 d 1 + x +
n2 d22 + (d x)2
dx
dx
dx
1
= n1 (d21 + x2 ) 2 · 2x + n2 (d22 + (d x)2 )
)
n1
x
d x
= n2
l1
l2
from the diagram we have sin ✓1 =
x)2
(3)
(4)
1
2
· 2(d
x) · ( 1) = 0
(5)
(6)
x
,
l1
sin ✓2 =
2
d x
.
l2
So n1 sin ✓1 = n2 sin ✓2
Figure 3: Problem 2
Problem 3: Ideal Prism
The incident ray is perpendicular to first face, so the light doesn‘t change
direction.
According to Snell‘s law
nprism sin ✓0 = sin ✓1
(7)
Incidence angle ✓1 = 10 , transmitted light angle ✓ = ✓1 ✓0 .
Using MATLAB, we can calculate ✓ vs the index nprism .(fig5)
The maximum transmitted light angle is 80 , with index nprism = 5.7588. If
index is larger, a total interval reflection will occur.
Figure 4: Problem 3
3
Figure 5: MATLAB Result
Problem 4: Surface reflections
From Wikipedia, I found the index of refraction of diamond is 2.419, water
is 1.333. So reflected intensity is
(
2.419 1.333 2
) = 8.378%
2.419 + 1.333
(8)
Note: Any reasonable index is allowed.
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Problem 5: Fluorescent slabs
First we calculate the angle where total interval reflection first occurs:
ns sin ✓c = na = 1
sin ✓c = 0.2857
(9)
)
cos ✓c = 0.9583
(10)
for the part of light that can‘t be totally internal reflected, every time it hits
the boundary, some portion transmits, others reflects. As the dist is infinite
in diameter, this part will eventually fully tranmit outside. For the part total
reflected, it will keep totally reflected, can‘t get out.
So solid angle of a cone of half-angle ✓c is 2⇡(1 cos ✓c ). Thus the fraction
of light evergy escapes from the top surface is:
2⇡(1
cos ✓c )
= 2.08%
4⇡
(11)
Figure 6: Problem 5
Textbook question
The last word is ’optics’.
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Problem 6: Slab waveguides
(A) For air nair = 1
According to Snell‘s law: nair sin ✓1 = ncone sin ✓2 with ✓1 = 45 and ncore =
1.5. So ✓2 = 28.13
✓3 =
⇡
2
✓2 = 61.87
(12)
Cladding here must ensure light is totally reflected on the boundary of
cladding and core. So maximum cladding index corresponds to critical angle
of ✓3
ncore sin ✓3 = nc ladding
)
max
ncladding = 1.323
(13)
The left and right side are symmetric. Therefore, light incident on face 2
reveals the same answer.
Figure 7: Problem 6
Figure 8: Problem 6
(B) Here I call the incident angle between core and cladding ✓30 . As we can
see from the figure, ✓30 is smaller than previous ✓3 .
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For critical angle situation, ncore sin ✓30 = ncladding . We should expect smaller
maximum cladding index. So some cladding index calculated in (A) cannot
ensure all the light propagate to the output face 2.
Here I call the incident angle between core and cladding ✓300 . As we can
see from the figure, ✓300 is smaller than previous ✓3 .
For critical angle situation, ncore sin ✓30 = ncladding . We should expect higher
maximum cladding index, thus ensures total internal reflection of first contact on the boundary.
Note: When two reflection surface are not parallel, the angle of incident
is a arithmetic sequence with relationship showed in fig9. ✓2 = ✓1 ✓0 .
So furthur incident from surface 1 keeps decrease, we can not make sure
there will be total internal incident every time. Furher incident angle from
surface 2 keeps grows, so total internal reflection every time.
Figure 9: Relationship of incident angle
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