snells law vector form
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Vector approach to ray tracing for reflection andrefraction
Y Kanoria and A G Parameswaran
Homi Bhabha Centre for Science Education, Tata Institute of Fundamental Research,
Mankhurd, Mumbai-400088, India.
E-mail: [email protected]
Abstract. The paper generalizes the vector approach to thin spherical lens ray
tracing given by Gatland (Gatland I R 2002 “Thin lens ray tracing”, American J.
Phys. 70(12) p 1184–6) to a general thin lens. A general result relating the incident
ray, the emergent ray and the nature of the lens is derived. This is then used to
analyze the behaviour of a cylindrical/sphero-cylindrical lens in a straightforward way.
A similar vector approach is also applied to reflection.
1. Introduction
Gatland[1] has given an elegant vector approach for thin spherical lens ray tracing
leading to the standard formulae for this case. In this paper, we have substantially
generalized the method and applied it to ray tracing for refraction by a general thinlens.
We obtain a general thin lens equation(28) that gives the direction of the emergent
ray in terms of the direction of the incident ray and the nature of the lens. The
assumptions under which the equation is derived are satisfied by most lenses used in
practice and this makes it widely applicable. We use this result to arrive at the formulae
for refraction by a spherical thin lens as an illustration. The application to refraction by
a sphero-cylindrical lens clearly illustrates the power of this approach. We also derive a
vector formulation of reflection and apply it to the case of a spherical mirror.
The vector approach, even for the elementary case of reflection and refraction by aspherical surface, is superior to the standard textbook approach [2], since in the latter
one must derive the formulae separately for convex and concave surfaces. Besides, the
standard derivations do not account for skew rays(rays not in the plane of the object and
principal axis), which must be considered to prove image formation. In the approach
presented, a single analysis takes care of all cases and also of skew rays.
We shall follow the New Cartesian sign convention.
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Vector approach to ray tracing for reflection and refraction 2
n
r
i
n1
NORMAL
P
2
s
SURFACE
REFLECTING
INCIDENT RAY
REFLECTED RAY
Figure 1. Vector formulation of reflection
2. Vector formulation of Reflection
Consider a ray incident at a point P on the surface of a mirror. Let n1 and n2 be the unit
vectors along directions of the incident and reflected ray respectively, with s being the
unit outward normal at P. Furthur, let ( n1) and ( n1)⊥ denote the component vectors
of n1 parallel and perpendicular to the surface at P respectively. Similarly, define ( n2)and ( n2)⊥. So,
n1 = ( n1) + ( n1)⊥ (1a )
n2 = ( n2) + ( n2)⊥ (1b)
Clearly, from the laws of reflection and equation (1a ) and (1b) we have,
( n1)⊥ = −( n2)⊥ = ( n1.s)s (2)
Also,
( n1) = ( n2) (3)
Equations (1a ), (1b), (2) and (3) may be combined to arrive at
n2 = n1 − 2( n1.s)s (4)
This equation encapsulates the two laws of reflection.
In the paraxial approximation,
n1.s ≈ −1 (5)
Equations (4) and (5) give
n2 = n1 + 2s (6)
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Vector approach to ray tracing for reflection and refraction 3
h
u = uk + h
u
Rk
r
O
OBJECT
P
SPHERICAL REFLECTING
MIRROR
+z
Figure 2. Image formation by spherical mirror
3. Analysis of image formation by spherical mirror
Let the origin be at the pole of the mirror and the principal axis be the z axis with
incident rays along the +z direction.Let k be the unit vector along +z. Consider a point
object situated at
ru = u k + h (7)
where h is perpendicular to the z axis.u is then object distance as per our convention.
Let r be the position vector of any point P on the mirror where a ray is incident.
Suppose u < 0 as shown in the figure. Then the incident ray is in the direction of r− ru.
We have,
n1 =(r
− ru)
|(r − ru)|≈ (r − ru)
(−u)
Substituting equation (7), we obtain
n1 = k +( h − r)
u(8)
We have assumed that r is close to 0 so that (r − ru) is almost parallel to the z axis.
For u > 0,
n1 =( ru
−r)
|( ru − r)|≈ ( ru − r)
u
Hence, equation (8) holds for u > 0 also.
For R < 0 as shown,
s =(R k − r)
|(R k − r)|
≈ (R k − r)
(−
R)
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Vector approach to ray tracing for reflection and refraction 4
= − k + r/R (9)
For R > 0,
s =
(r
−R k)
|(r − R k)| ≈(r
−R k)
R
Equation (9) thus holds for R > 0 also.
From equations (6), (8) and (9),
n2 = − k + h
u+ r(
2
R− 1
u) (10)
Any point on the reflected ray is given by
p(α) = r + α n2 (11)
α is positive for real points and negative for virtual points. From equations (10) and
(11)
p(α) = r[1 + α(2
R− 1
u)] + α(
h
u− k) (12)
To obtain the location of the image rv, we obtain the value of α for which p(α) is
independent of r. This value is given by
α = α0 =−1
2R
− 1u
(13)
Equations (12) and (13) give
p(α0) =
α0
u
h−
α0
k (14)
Let
rv = v k + m h (15)
where v is image distance and m can be interpreted as magnification. Comparing
equations (14) and (15),
v = −α0 and m = α0/u (16)
Equations (13) and (16) give
1
v+
1
u=
2
Rand m =
−v
uwhich are the standard formulae for reflection by a spherical mirror.
4. Vector formulation of refraction
Define a ray vector corresponding to a light ray to be a vector along the direction of
the ray, having a magnitude equal to the refractive index of the medium. Consider a
ray travelling in a medium of refractive index µ1 incident at a point P at the interface
of the medium with another medium of refractive index µ2. The notation in this case
is: n1 is the ray vector describing the incident ray, n2 is the ray vector describing the
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Vector approach to ray tracing for reflection and refraction 5
n
n
i
r
2
NORMAL
INCIDENT RAY
REFRACTED RAY
REFRACTING SURFACE
1
sP
Figure 3. Vector formulation of refraction
refracted ray, s is the unit vector along the normal in the direction of light, and ( n1),
( n1)⊥, ( n2), ( n2)⊥ are the vector components of n1 and n2 defined as before. Using
Snell’s law, we arrive at
( n2)⊥ = (
(µ2)2 − (µ1)2 + ( n1.s)2)s (17)
The laws of refraction give us
( n1) = ( n2) (18)
Equations (17) and (18) may be combined to arrive at
n2 = n1 − ( n1.s)s + (
(µ2)2 − (µ1)2 + ( n1.s)2)s
This expression looks quite unwieldy until we assume that i is small so that n1.s ≈ µ1.
Under this approximation we get
n2 = n1 + (µ2 − µ1)s (19)
which is the basic equation given by Gatland[1].
Using equation (19), and following exactly the same procedure as in section 3, we
arrive at the formulae for refraction at a single spherical surface
µ2
v −
µ1
u
=µ2 − µ1
R
and m =µ1v
µ2u
(20)
These expressions can be used to easily derive 1f
= 1v− 1
uand m = v
ufor a thin lens with
spherical surfaces[3].
5. Analysis of deviation of a ray by a general thin lens
Consider a thin lens of constant refractive index µ. Suppose both bounding surfaces are
almost perpendicular to the z axis. Take some x and y axis forming an right-handed
orthonormal system with z axis. Let the bounding surfaces be:
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Vector approach to ray tracing for reflection and refraction 6
n
G
n n2
s2
s1
INCIDENT RAY
F
NORMAL TO F
NORMAL TO G
1
EMERGENT
RAY
Figure 4. A General thin lens
F: z = f (x, y) and G: z = g(x, y)Define
t(x, y) = g(x, y) − f (x, y) (21)
where t is the thickness of the lens. Consider an incident ray travelling along the +z
direction. Let n1, n and n2 be the ray vectors describing the incident ray, the ray within
the lens and the emergent ray respectively. Let s1 and s2 be the unit normals to the
surfaces F and G respectively, at points along the path of the ray.
Since s1 is the normal to the surface F,
s1 = (z
−f (x, y))
|(z − f (x, y))|=
−f x i − f y j + k 1 + f 2x + f 2y
Since the bounding surface F is almost perpendicular to k, f 2x + f 2y << 1, so
s1 ≈ −f x i − f y j + k (22)
Similarly,
s2 ≈ −gx i − gy j + k (23)
We assume that n1 is almost parallel to k. Since s1 is also almost parallel to k, thismeans the angle of incidence is small and equation (19) is applicable for the appropriate
variables.
n = n1 + (µ − 1) s1 (24)
n2 = n + (1 − µ) s2 (25)
Eliminating n, we get
n2 = n1 + (µ − 1)( s1 − s2) (26)
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Vector approach to ray tracing for reflection and refraction 7
C CR
R
G
Q
F
P
0
REFRACTED
RAY
1
1
2
2+z
INCIDENT
RAY
Figure 5. Lens with spherical surfaces
From equations (21), (22) and (23)
s1 − s2 = (gx − f x) i + (gy − f y) j
= tx i + ty j
= t(x, y) (27)
Using (26) and (27),
n2 = n1 + (µ − 1)t (28)
This expression gives the direction of the emergent ray directly, given the incidentray and the nature of the lens. An interesting point to note is that under the given
assumptions, the lens is characterized only by its thickness, not the nature of the
individual surfaces. This conclusion and equation (28) follow also from the use of
Fermat’s principle. Equation (28) immediately shows that light bends in the direction
of increasing thickness of the lens.
The use of equation (28) can be demonstrated for the simple case of a thin lens
with spherical surfaces, which has already been discussed in the previous section. The
principal axes is taken as the z axis and the origin as its intersection with the surface
F. Let r =√
x2 + y2. Let r be the position vector of the point P on the lens, where the
incident ray falls. For a thin lens, this is nearly equal to the position vector of the pointQ where the emergent ray leaves the lens. Let r̂ be the unit vector along the projection
of OP in the xy plane. In the paraxial approximation, we have
r ≈ rr̂ (29)
Note our definition of r, r and r̂.
A little elementary geometry gives
t = R2(1 −
1 − r2
R22
) − R1(1 −
1 − r2
R21
) + c (30)
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Vector approach to ray tracing for reflection and refraction 8
where c = t(0, 0). Note that the signs have been taken care of. Assume r << R1 and
r << R2 (paraxial approximation). Using binomial expansion in equation (30),
t
≈R2(
r2
2R2
2
)
−R1(
r2
2R2
1
) + c
=r2
2(
1
R2
− 1
R1) + c (31a )
t = (1
R2
− 1
R1
)rr̂ (31b)
Using equations (29) and (31b), we obtain
t ≈ (1
R2
− 1
R1
)r (32)
From equations (28) and (32), we have:
n2 = n1 + (µ − 1)[(1
R2−
1
R1)r] (33)
Define f by
1
f = (µ − 1)(
1
R1
− 1
R2) (34)
From equations (33) and (34), we have,
n2 = n1 − r
f (35)
Take a point object at ru = u k + h. As before,
n1 = k + h − ru
(36)
So, from equations (35) and (36),
n2 = k + h
u− r(
1
u+
1
f ) (37)
Any point on the emergent ray is
p = r + α n2
= r(1 − α(1
u+
1
f )) + α( k +
h
u)
Proceeding as before, we arrive at
1
v− 1
u=
1
f and m =
v
u(38)
These along with equation (34) are the standard formulae for a thin lens with spherical
surfaces.
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Vector approach to ray tracing for reflection and refraction 9
r
O
ru
OBJECT
SURFACE
SPHERICAL
SURFACE
CYLINDRICAL
+x
+z
Figure 6. Side View of a Sphero-cylindrical Lens
r
+y
Oru
SURFACE
SPHERICAL
SURFACE
CYLINDRICAL
OBJECT
+z
Figure 7. Top View of a Sphero-cylindrical Lens
6. Refraction by Sphero-Cylindrical Lens
Consider a thin sphero-cylindrical lens, with the z axis being the principal axis. Let the
surface to the left be spherical with radius of curvature Rs and the surface to right be
cylindrical with radius of curvature Rc, the axis of the cylinder being along the x axis
(refer to figure 6,7). Consider a point object at ru = u k. Define r, r̂ and r as in the
derivation of deviation of a ray by a thin lens with spherical surfaces. Equation (29)
holds since the lens is thin. We have,
n1 = k − r
u(39)
Let
rr̂ = x0 i + y0 j (40)
The figures are drawn for the case u < 0, Rs > 0, Rc < 0 i.e. for a real object and both
surfaces convergent. However, it is easy to see that the analysis which follows is general.
Note how the formation of an astigmatic pencil and the change in its cross section with
the z coordinate can be derived without any complicated arguments or mathematics.
We must find the thickness of the lens t as a function of x and y. The equation of
the spherical surface is,
z = Rs(1 −
1 − r2
R2s
) − c1 (41)
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Vector approach to ray tracing for reflection and refraction 10
The equation of the cylindrical surface is
z = Rc(1 −
1 − y2
R2c
) + c2 (42)
c1 and c2 are small positive constants. Note that the signs have been taken care of.Assume r << Rs and r << Rc (paraxial approximation). Using the binomial expansion,
we obtain,
t ≈ − r2
2Rs
+y2
2Rc
+ c (43)
where c = t(0, 0) = c1 + c2.
t ≈ (− r
Rs
+y
Rc
j) (44)
From equations (28), (39) and (44), we have
n2 = k − ru
+ (µ − 1)(− rRs
+ yRc
j) (45)
Any point on the emergent ray is
p = r + α n2 (46a )
= x i + y i + z k (46b)
where
x = x0(1 + α(−1
u− µ − 1
Rs
)) (47a )
y = y0(1 + α(
−1
u+ (µ
−1)(
−1
Rs
+1
Rc
))) (47b)
z = α (47c)
Equations (46b), (45) and (40) have been used.
Suppose Rs → ∞. The lens is cylindrical(or plano-cylindrical). Consider α = α0
such that1
α0− 1
u= −µ − 1
Rc
(48)
Then we have
y = 0 (49)
z = α0 (50)
x = x0α0(1
α0− 1
u− 0) = −x0α0(µ − 1)
Rc
(51)
Thus we have a focal line at this position, parallel to the axis of the cylinder. At any
value of α (i.e. of z) x depends only on x0, y depends on y0. Hence, the section of the
emergent beam at any position along z is rectangular if the plane face of the lens is
rectangular. These are the results quoted in [4].
Now suppose Rs is finite. For α = z = α1 such that
1
α1 −
1
u
=µ − 1
Rs
(52)
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Vector approach to ray tracing for reflection and refraction 11
we have x = 0 i.e. the beam converges to a horizontal line. For α = z = α2 such that
1
α2− 1
u= (µ − 1)(
1
Rs
− 1
Rc
) (53)
we have y = 0 i.e. the beam converges to a vertical line. If a section of the lens normalto the z axis is circular with radius R, we have
x20 + y2
0 ≤ R20 (54)
Using equations (47a ) and (47b), we obtain
x2
(1 + α(− 1u
− µ−1Rs
))2+
y2
(1 + α(− 1u
+ µ−1(− 1
Rs+ 1
Rc)))2
≤ R20 (55)
Clearly, the cross-section of the beam is elliptical. Thus we are able to explain the
formation of the astigmatic pencil with all the quantitative details. The results obtained
match the standard results[4].
7. Conclusion
The general thin lens equation (28) is an important and useful result. The analysis
leading to it can be profitably incorporated in undergraduate physics texts. Further,
equation (28) can be used for the analysis of different kinds of lenses and even to find
the nature of a lens which bends light in a desired way.
References
[1] Gatland I R 2002 Thin lens ray tracing, American J. Phys. 70(12) p 1184–6
[2] Smith C J 1960 A Degree Physics, Part 3 (Optics) (London: Edward Arnold) p 105–10 p 125–33
[3] Smith C J 1960 A Degree Physics, Part 3 (Optics) (London: Edward Arnold) p 137–9
[4] Smith C J 1960 A Degree Physics, Part 3 (Optics) (London: Edward Arnold) p 230–6
Acknowledgments
This work was done as part of the National Initiative for Undergraduate Science,
undertaken at the Homi Bhabha Centre for Science Education, Tata Institute of
Fundamental Research. It is a pleasure to thank Prof. Arvind Kumar for his help
and advice.
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Vector approach to ray tracing for reflection and refraction 12
Images
• Figure 1: Vector formulation of reflection
•Figure 2: Image formation by spherical mirror
• Figure 3: Vector formulation of refraction
• Figure 4: A General thin lens
• Figure 5: Lens with spherical surfaces
• Figure 6: Side View of a Sphero-cylindrical Lens
• Figure 7: Top View of a Sphero-cylindrical Lens