snells law vector form

8/2/2019 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.



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)




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)




= − 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)


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)


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




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:




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)




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)




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.




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)




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)




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.




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

snells law vector form

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