Chapter

Geometric Optics

Blinn College - Physics 1402 - Terry Honan

. - Wave Fronts, Rays and Reflection

Wave Fronts and Rays

There are two ways we can visualize waves, as wave fronts or as rays.  If we mark the crest of each wave then these form the wave fronts. For a plane wave the crest corresponds to parallel planes moving at the wave speed perpendicular to the plane. Light from a point source is a spherical wave. A spherical wave from a distant source approaches a plane wave. In both cases, the distance between wave fronts is one wavelength and the rays are perpendicular to the wave fronts. A light ray is what you expect it to be, it points in the direction of the light propagation.

Reflection of Light

When we describe the angle that light makes with a surface, we measure the angle of the ray from the normal to the surface. When light reflects off a smooth shiny surface, it reflects so that the incident angle equals the reflected angle. This is known as the law of reflection. Most of the light we see is reflected light. Reflection off shiny smooth surfaces is called specular reflection. Most surfaces are rough on the order of the wavelengths of light. Because the normal to a rough surface is varying, the light then reflects in random directions. This is called diffuse reflection. Most of the light that we see is from the diffuse reflection off surfaces.

. - Refraction

Light in a Medium

When we refer to c as the speed of light, it is implied that it is the speed in a vacuum. In a medium light slows down. It slows down by a factor called the index of refraction; this is a material dependent constant n defined by

v=(c/n),

where v is the speed in the medium.  Clearly n≥1 and the equality applies to a vacuum.

When light hits an interface between indices (Subscript[n, 1]) and (Subscript[n, 2]) both the frequency and wavelength cannot remain unchanged, since

f λ=v=(c/n).

The frequency is the same on both sides of the interface.  This is easy to see.  At the interface the incoming radiation is varying at some frequency and generally driving a system at some frequency induces oscillations at the same frequency.  Given that the frequencies are equal

f=(Subscript[f, 1])=(Subscript[f, 2])

the wavelengths must change.  Since wavelength is proportional to the wave speed we get

((Subscript[λ, 2])/(Subscript[λ, 1]))=((Subscript[v, 2])/(Subscript[v, 1]))=((Subscript[n, 1])/(Subscript[n, 2]))   ⟹   (Subscript[n, 1]) (Subscript[λ, 1])=(Subscript[n, 2]) (Subscript[λ, 2]).

To view this a different way, if we define the vacuum wavelength as  (Subscript[λ, 0])=c/f  then the wavelength in a medium with index n is

λ=((Subscript[λ, 0])/n).

The index of refraction is a property of a material. Because it is a ratio of a speed to a speed, it is dimensionless; it has units. For a vacuum it is exactly 1. For air it is near 1 and we will usually take it as one, unless directed otherwise.

({{Material , Index of Refraction - n}, {Vacuum, 1  (exact)}, {Air (0°C, 1 atm), 1.000277}, {Water, 1.333}, {Plate Glass, 1.52}, {Diamond, 2.417}, {Polycarbonate, 1.60}})

Snell's Law

As a consequence of the speed of light changing across an interface light will bend as it hits the interface. This is called refraction. Consider light passing from one medium to another through a flat interface.  Take it to move from medium 1 with index (Subscript[n, 1]) to medium 2 with index (Subscript[n, 2]).  We will, as usual, measure the angles of the rays relative to the normal to the surface.

The diagram shows wave fronts and rays on either side of the interface.  Since the rays are perpendicular to the wave fronts, the angle between the rays and the normal are the same as the angle between the wave fronts and the interface.  Enlarging the triangles at the center

and labeling the common side as D gives

sin (Subscript[θ, 1])=((Subscript[λ, 1])/D) and  sin (Subscript[θ, 2])=((Subscript[λ, 2])/D)  ⟹  ((sin (Subscript[θ, 2]))/(sin (Subscript[θ, 1])))=((Subscript[λ, 2])/(Subscript[λ, 1]))=((Subscript[n, 1])/(Subscript[n, 2])).

We can now write the law of refraction, called Snell's law, as

(Subscript[n, 1]) sin (Subscript[θ, 1])=(Subscript[n, 2]) sin (Subscript[θ, 2]).

The usual convention with Snell’s law is that the light moves from medium 1 to medium 2.

Incident, Reflected and Refracted Rays

When light in a medium with index (Subscript[n, 1]) hits a medium with index (Subscript[n, 1]) some of the light passes through and the angle (Subscript[θ, 2]) is found by Snell’s law. In addition to refraction some of the light is reflected back by the law of reflection. The incoming ray is called the incident ray.

Note that in this example the angle in medium 2 is larger than in medium 1.  This implies, by Snell's law, that the index in 2 is smaller than in 1.

(Subscript[θ, 1])<(Subscript[θ, 2])    ⟹  sin (Subscript[θ, 1])<sin (Subscript[θ, 2])    ⟹    (Subscript[n, 1])>(Subscript[n, 2])

Example . - Light from Air to Glass

Solution

(Subscript[n, 1])=1 ,  (Subscript[n, 2])=1.52 ,  (Subscript[θ, 1])=35° and  (Subscript[θ, 2])=?

(Subscript[n, 1]) sin (Subscript[θ, 1])=(Subscript[n, 2]) sin (Subscript[θ, 2])  ⟹  (Subscript[θ, 2])=(sin^(-1))(((Subscript[n, 1])/(Subscript[n, 2]))sin (Subscript[θ, 1]))=22.2°

Solution

(Subscript[λ, 0])=632.9(QuantityUnit)=632.9×(10^(-9))(QuantityUnit) , n=1.52  and  c=3.00×(10^8)(QuantityUnit)/(QuantityUnit)

v=c/n=1.97×(10^8)(((QuantityUnit))/((QuantityUnit)))

Solution

λ=((Subscript[λ, 0])/n)=416(QuantityUnit)=4.16×(10^(-7))(QuantityUnit)

Solution

f=(Subscript[f, 0])=(c/(Subscript[λ, 0]))=4.74×(10^14)(QuantityUnit)

Total Internal Reflection

Consider the case where light moves across an interface to a region with lower index.

(Subscript[n, 1])>(Subscript[n, 2])

It follows that the angle from the normal increases ((Subscript[θ, 1])<(Subscript[θ, 2])).  Moreover, there is some angle where there is no refracted ray.  In this case there is only the reflected ray.  This is what we call total internal reflection.  This occurs at incident angles above some critical angle

(Subscript[θ, 1])≥(Subscript[θ, crit]),

where the critical angle occurs when the refracted angle goes to 90°.  That is, (Subscript[θ, 1])=(Subscript[θ, crit])  when  sin (Subscript[θ, 2])=1, or

sin (Subscript[θ, crit])=((Subscript[n, 2])/(Subscript[n, 1])).

Example . - A Water to Air Interface

Solution

(Subscript[n, 1])=1.333 ,  (Subscript[n, 2])=1, sin (Subscript[θ, crit])=((Subscript[n, 2])/(Subscript[n, 1]))  ⟹  (Subscript[θ, crit])=(sin^(-1))(1/1.333)=48.6°

Solution

⟹    tan (Subscript[θ, 1])=(x/d)   where d=1.5(QuantityUnit)

x=d tan (Subscript[θ, 1])   ⟹   (Subscript[x, max])=d tan (Subscript[θ, crit])=d tan 48.6°=1.70(QuantityUnit)

Prisms and Dispersion

Deflection by a prism

Consider light hitting a prism as shown above.  The apex angle Φ is the angle between the two refracting surfaces in the prism.  If the ray incident on the prism is at an angle (Subscript[θ, 1])from the normal it will refract to an angle (Subscript[θ, 2]), which can be found by Snell’s law.

sin (Subscript[θ, 1])=n sin (Subscript[θ, 2])

Now consider the triangle formed by the ray inside the prism and the top of the prism.  Summing the internal angles of this triangle must give 180°.

(90°-(Subscript[θ, 2]))+(90°-(Subscript[θ, 2]^′))+Φ=180°     ⟹    (Subscript[θ, 2])+(Subscript[θ, 2]^′)=Φ

The above expression allows us to find (Subscript[θ, 2]^′) from (Subscript[θ, 2]) and Φ.  This lets us find (Subscript[θ, 1]^′).

sin (Subscript[θ, 1]^′)=n sin (Subscript[θ, 2]^′)

The total angle of deflection δ is found by summing the bending at both interfaces.  At the first the ray bends by (Subscript[θ, 1])-(Subscript[θ, 2]) and at the second by  (Subscript[θ, 1]^′)-(Subscript[θ, 2]^′).  It follows that δ is the sum.

δ= ((Subscript[θ, 1])-(Subscript[θ, 2]))+((Subscript[θ, 1]^′)-(Subscript[θ, 2]^′))=(Subscript[θ, 1])+(Subscript[θ, 1]^′)-Φ

Dispersion

A prism splits light into the visible spectrum: red, orange, yellow, green, blue, indigo and violet.  We have just seen that using Snell's law and knowing the prism’s apex angle Φ and the incident angle (Subscript[θ, 1]), we can uniquely solve for the position of the ray leaving the prism in terms of the index of refraction of the prism.  The fact that different colors refract to different angles implies that the index of refraction varies with color (wavelength).  This is known as dispersion.

In a prism, red is the color that refracts the least; this means the index of red is the least.  Red has the longest visible wavelength.  It is generally the case that the index of refraction decreases with wavelength.  For instance in the case of crown glass the tabulated index is 1.52; this is a mean value.  As the wavelength increases from 400 to 700 nm, the index decreases from about 1.53 to 1.51.

. - Images from Spherical and Parabolic Mirrors

Concave Mirrors

Spherical Mirrors and Spherical Aberration

When light from infinity (a plane wave) parallel to the central axis of a concave spherical mirror reflects off the mirror, the rays near the central axis converge toward a point called the focal point, which we will see is at a position half the radius from the mirror.

The rays far from the central axis miss the focal point.  This is called spherical aberration.

Parabolic Mirrors

The geometrical shape that causes all rays to converge to a single focal point is a paraboloid, which is a parabola under rotation.

As long as the distance from the central axis is small (compared to the sphere's radius) the sphere approximates a paraboloid well.  This is the basic assumption of our analysis of spherical mirrors.

Plane Waves from off the Central Axis

Point Source off the Central Axis

Images from Concave Mirrors

We need to mathematically relate the image position (Subscript[d, i]) and image height (Subscript[h, i]) to the object position (Subscript[d, o]) and object height (Subscript[h, o]) for the case of a concave spherical mirror.  In the preceding diagram it is clear that every ray from the tip of the object converges to the tip of the image.  It suffices to draw just two rays to find the image position.  One ray we will draw is from the tip of the object to the point where the central axis hits the mirror; this will reflect back at an equal angle below the central axis.  The other ray we will consider passes through the center of the sphere; this will hit the surface normally and reflect straight backward.

These two rays give two pairs of similar triangles.

These give the expressions

((|(Subscript[h, i])|)/(Subscript[h, o]))=((Subscript[d, i])/(Subscript[d, o]))  and  ((|(Subscript[h, i])|)/(Subscript[h, o]))=((R-(Subscript[d, i]))/((Subscript[d, o])-R))

((Subscript[d, i])/(Subscript[d, o]))=((R-(Subscript[d, i]))/((Subscript[d, o])-R))   ⟹   (1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(2/R)

If the object is at infinity ((Subscript[d, o])->∞) the image is at the focal point ((Subscript[d, i])=f).  This gives f=R/2.  Since the image is inverted we choose, by convention, that (Subscript[h, i])<0 and thus |(Subscript[h, i])|=-(Subscript[h, i]).  We can rewrite the above expressions as

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f)   and   m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o])).

These expressions will apply generally to spherical mirrors, either concave or convex, and to thin lenses, either converging or diverging.  m is defined as the magnification; it has the same sign convention as the image height (Subscript[h, i]).

The preceding image is called a real image.  A real image occurs when the light rays converge to a point. A real image is on the side where the light is. If you hold a screen to a real image you have a projected image on the screen.  If the object is inside the focal point, it turns out that the reflected rays do not converge.  We trace the same rays as before but to find the image we trace the diverging rays backward to see where the reflected light rays appear to originate; this is the image position.  Mathematically, when the object is inside the focal point  0<(Subscript[d, o])<f  the image position is negative (Subscript[d, i])<0. There is no light at a virtual image. We will see that a plane mirror, like a bathroom mirror, gives only virtual imagesl that are behind the mirror; ther eis no light there.

Example . - Concave Spherical Mirror

Solution

R=60(QuantityUnit)  ⟹  f=(R/2)=30(QuantityUnit) ,  (Subscript[h, o])=10(QuantityUnit)  and  (Subscript[d, o])=40(QuantityUnit)

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f) ⟹ (Subscript[d, i])= ((((1/f)-(1/(Subscript[d, o]))))^(-1))=120(QuantityUnit)

m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o]))  ⟹   (Subscript[h, i])=-(Subscript[h, o])((Subscript[d, i])/(Subscript[d, o]))=-30(QuantityUnit)   and   m=-((Subscript[d, i])/(Subscript[d, o]))=-3

With as single mirror (or lens as we will see later) a real image is always inverted and inverted images are always real. Inverted images have negative (Subscript[h, i]) and m.

We can use any length unit (cm, m, ft, in, ...) as long as we are consistent.

Solution

R=60(QuantityUnit)  ⟹  f=(R/2)=30(QuantityUnit) ,  (Subscript[h, o])=10(QuantityUnit)  and  (Subscript[d, o])=20(QuantityUnit)

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f) ⟹ (Subscript[d, i])= ((((1/f)-(1/(Subscript[d, o]))))^(-1))=-60(QuantityUnit)

m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o]))  ⟹   (Subscript[h, i])=-(Subscript[h, o])((Subscript[d, i])/(Subscript[d, o]))=+30(QuantityUnit)   and   m=-((Subscript[d, i])/(Subscript[d, o]))=+3

Images from Convex Mirrors

The center of a convex mirror is behind the mirror.  We take the radius to be negative.  An incident plane wave parallel to the central axis will reflect away from a focal point behind the mirror.  When the reflected rays are extrapolated backward they meet at the focal point.  We also take the focal length to be negative.

Light from an off-axis point source will reflect away from a point behind the mirror, the image position.  This will create a virtual image.

The ray tracing is similar to the concave case except that the center is now behind the mirror.

An analogous analysis with similar triangles yields the same expressions

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f)   and   m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o])).

It is still true that f=R/2 but now both R and f are negative.  It follows that since (Subscript[d, o])>0 we will always get (Subscript[d, i])<0 and thus a virtual image.

Example . - Convex Spherical Mirror

Solution

R=-60(QuantityUnit)  ⟹  f=(R/2)=-30(QuantityUnit) ,  (Subscript[h, o])=10(QuantityUnit)  and  (Subscript[d, o])=20(QuantityUnit)

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f) ⟹ (Subscript[d, i])= ((((1/f)-(1/(Subscript[d, o]))))^(-1))=-12(QuantityUnit)

m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o]))  ⟹   (Subscript[h, i])=-(Subscript[h, o])((Subscript[d, i])/(Subscript[d, o]))=+6(QuantityUnit)   and   m=-((Subscript[d, i])/(Subscript[d, o]))=+0.6

Flat Mirrors

A flat mirror can be viewed as a special case of a spherical mirror with R->±∞.  This implies that f->±∞  and  1/f->0

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f)=0   ⟹  (Subscript[d, i])=-(Subscript[d, o])

m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o]))  ⟹  m=1 and  (Subscript[h, i])=(Subscript[h, o])

This is what we would expect:  There is an upright virtual image the same size as the object and equal distance behind the mirror.

Example . - A Make-up Mirror

Solution

m=3,  (Subscript[d, 0])=10(QuantityUnit) ,  R=?

m=-((Subscript[d, i])/(Subscript[d, o]))  ⟹   (Subscript[d, i])=-m (Subscript[d, o])=-30(QuantityUnit)

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f) ⟹ f= ((((1/(Subscript[d, o]))+(1/(Subscript[d, i]))))^(-1))=15(QuantityUnit)

f=(R/2)   ⟹   R=2f=30(QuantityUnit)

. - Images from Thin Lenses

Thin Lenses

A lens has two refracting surfaces. Solving for images from two refracting surfaces is a complex thing but when the lens is thin, meaning that the two surfaces are close to each other, the problem becomes simple. To relate the image position and height to the object position and height, we have the same equations we used for mirrors.

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f)   and   m=((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o])).

Although we have two surfaces that refract we will treat a thin lens as if all refraction happens at the central plane

General Sign Conventions in Optics

In the case above where images were formed by mirrors, the sign conventions were relatively simple.  (Subscript[d, o]) is positive when the object is on the front side of the mirror.  Similarly, (Subscript[d, i]), f and R are positive when the image, focal point and center, respectively, are on the mirror's front side.

With a lens the light leaves on the opposite side where it enters.  Label the ray that approaches a mirror or lens as the entering ray and the ray leaving as the exiting ray; this is the reflected ray for mirrors or the refracted ray for a lens.

A general sign convention can now be given.  (Subscript[d, o]) is positive when the object is on the side of the entering ray, the side where the light originates.   (Subscript[d, i]), f and R are positive when the image, focal point and center, respectively, are on the side of the exiting ray, on the side where the light ends up.

Converging Lenses

A thin lens with a positive focal length is called a converging lens.  This is analogous to a concave mirror mathematically.  If the object distance is larger than the focal length (Subscript[d, o])>f  then there is a real image and when the object distance is less than the focal length (Subscript[d, o])<f there is a virtual image.  To trace the rays: Draw one ray that passes straight through where the central axis hits the lens.  Draw the other ray parallel to the central axis; this will bend through the focal point.  The two rays (as all rays do) will converge to the image.  Below are examples of ray tracing in both cases.

Diverging Lenses

A diverging lens has a negative focal length.  The ray tracing is similar but the ray that begins parallel to the central axis will diverge away from the focal point, which is on the same side as the object.  Here is the ray tracing for a diverging lens.

Example . - A Slide Projector

Solution

f=12(QuantityUnit) ,  (Subscript[h, o])=3.5(QuantityUnit)  and  (Subscript[h, i])=±210(QuantityUnit).

((Subscript[h, i])/(Subscript[h, o]))=-((Subscript[d, i])/(Subscript[d, o]))  ⟹  (Subscript[d, i])=-((Subscript[h, i])/(Subscript[h, o]))(Subscript[d, o])

(Subscript[h, i])=-210(QuantityUnit)   ⟹   (Subscript[d, i])=-((Subscript[h, i])/(Subscript[h, o]))(Subscript[d, o])=60 (Subscript[d, o])

(1/(Subscript[d, o]))+(1/(Subscript[d, i]))=(1/f)  ⟹  (1/(Subscript[d, o]))+(1/(60 (Subscript[d, o])))=(1/(12(QuantityUnit)))

12(QuantityUnit) (1+(1/60))=(Subscript[d, o])

(Subscript[d, o])=12.2(QuantityUnit)   and  (Subscript[d, i])=60 (Subscript[d, o])=732(QuantityUnit)