Chapter

Geometric Optics

Blinn College - Physics 2326 - Terry Honan

. - Properties of Light

The Speed of Light

The speed of light in a vacuum is approximately c≃3.0×(10^8) m/s.  Because of its most fundamental nature we can choose our units to make c have the exact value

c=2.99792458×(10^8) m/s.

The definition of a second is given in terms of our most accurate method of measuring time, with atomic clocks, and the above definition of c then gives a definition of the meter.

Historically, the first good estimate of the speed of light was made by Roemer.  By plotting the period of Io, one of Jupiter's moons, Roemer was able to explain the lack of periodicity in Io's orbit in terms of the time difference for light from Io to reach the Earth as the relative distance between Io and Earth varies.  An accurate measure of c can be found by shining a light through a rapidly rotating toothed wheel and reflecting it off a distant mirror.

The speed of light is a fundamental speed limit.  It is impossible to send any information faster than c.

Example . - Time of Travel for Light.

(Subscript[R, E,S])=1.50×(10^11)(QuantityUnit)

Solution

t=(d/c)=((Subscript[R, E,S])/c)=500(QuantityUnit)=8.33(QuantityUnit)

(Subscript[R, E,M])=3.84×(10^8)(QuantityUnit)

Solution

t=(d/c)=((Subscript[R, E,M])/c)=1.28(QuantityUnit)

Solution

t=(10^(-9))(QuantityUnit)  ⟹  d=c t=0.3(QuantityUnit)≃1 foot

Solution

1(QuantityUnit)=c×(1(QuantityUnit))=3.00×(10^8) (m/((QuantityUnit)))×365.24(QuantityUnit)×((24(QuantityUnit))/day)×((3600(QuantityUnit))/((QuantityUnit)))=9.47×(10^15)(QuantityUnit)

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.

The plane wave correspond to our plane wave solution where the fields are constant along planes; these planes are the wave fronts.  Light from a point source gives spherical waves. In both cases, the distance between wave fronts is a wavelength. When a point source is a long distance away the spherical wave can also be described as a plane wave. Light rays are what you would expect from the term; they are “beams” of light perpendicular to the wave front. The beam of a laser behaves as a ray.

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.

({{         Material          ,     Index of Refraction - n    }, {Vacuum, 1  (exact)}, {Air (0°C, 1 atm), 1.000277}, {Water, 1.333}, {Crown Glass, 1.52}, {Diamond, 2.417}, {Polycarbonate, 1.59}})

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.  The wavelength is then proportional to the wave speed and 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).

. - Refraction and Reflection

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 of Light and 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]).

Incident, Reflected and Refracted Rays

In addition to refraction some of the light is reflected.  The so-called law of reflection states that the angle, measured from the normal, of the reflected ray is the same as the incident.

Note that in this example the angle in medium 2 is larger than in medium 1.  Since sine is an increasing function then, by Snell's law, 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 Water

Solution

(Subscript[n, 1])=1,  (Subscript[n, 2])=1.333  and   (Subscript[θ, 1])=35°.

(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]))=25.5°

Solution

c=3.00×(10^8) (m/s)  and  n=1.333   ⟹   v=(c/n)=2.25×(10^8) (m/s)

Solution

(Subscript[λ, 0])=560(QuantityUnit)=560×(10^(-9))(QuantityUnit) .

λ=(Subscript[λ, 0])/n=420(QuantityUnit)=4.20×(10^(-7))(QuantityUnit)

Solution

f=(Subscript[f, 0])=(c/(Subscript[λ, 0]))=5.36×(10^14)(QuantityUnit)

Total Internal Reflection

Suppose a light ray originates under water at an angle of (Subscript[θ, 1]) from the normal to the surface and emerging at an angle (Subscript[θ, 2]) in air. We will take the index for air to be one.  How will the angle (Subscript[θ, 2]) change as we vary (Subscript[θ, 1])?  The index of refraction for water is (Subscript[n, water])=1.333.  Notice that (Subscript[θ, 2]) increases as (Subscript[θ, 1]) increases and is larger than (Subscript[θ, 1]).  There is a point when the angle (Subscript[θ, 2]) cannot be found; this is when the calculation of (sin^(-1))((Subscript[n, water])sin((Subscript[θ, 1]))) fails because you are evaluating the arcsin of an argument larger than one.

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

({{      (Subscript[θ, 1])      ,       (Subscript[θ, 2])      }, { 10°,  13.38°}, { 20°,  27.12°}, { 30°,  41.80°}, { 40°,  58.96°}, { 45°,  70.49°}, { 48°,  82.14°}, { 49°, -}})

Imagine standing in a pool of depth D with your eye just above the surface of water. If there is a small object on the floor of the pool a horizontal distance of x from your eye.

As you move backward to increase x there will be a point where the object disappears. What you see is the refracted ray and with a sufficiently large x that refracted ray will disappear. This is known as “total internal reflection”.

Consider the general 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 . - Critical Angle from Water to Air

Solution

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

Example . - An Object at the Bottom of a Pool

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.

Example . - Angular Separation of Colors in a Prism

Solution

Φ=25°,  (Subscript[θ, 1])=15°, (Subscript[n, red])=1.51  and  (Subscript[n, violet])=1.53

(Subscript[θ, 2,red])=(sin^(-1))((1/(Subscript[n, red]))sin (Subscript[θ, 1]))=9.869°  and  (Subscript[θ, 2,violet])=(sin^(-1))((1/(Subscript[n, violet]))sin (Subscript[θ, 1]))=9.739°

(Subscript[θ, 2,red]^′)=Φ-(Subscript[θ, 2,red])=15.131°  and  (Subscript[θ, 2,violet]^′)=Φ-(Subscript[θ, 2,violet])=15.261°

(Subscript[θ, 1,red]^′)=(sin^(-1))((Subscript[n, red]) sin (Subscript[θ, 2,red]^′))=23.212°  and  (Subscript[θ, 1,violet]^′)=(sin^(-1))((Subscript[n, violet]) sin (Subscript[θ, 2,violet]^′))=23.748°

(Subscript[θ, 1,violet]^′)-(Subscript[θ, 1,red]^′)=0.54°

. - 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 s' and image height h' to the object position s and object height h 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

((|h'|)/h)=((s')/s)  and  ((|h'|)/h)=((R-s')/(s-R))

((s')/s)=((R-s')/(s-R))   ⟹   (1/s)+(1/(s'))=(2/R)

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

(1/s)+(1/(s'))=(1/f)   and   m=((h')/h)=-((s')/s).

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 h'.

The preceding image is called a real image.  A real image occurs when the light rays converge to a point.  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<s<f  the image position is negative s'<0.

Example . - Concave Spherical Mirror

Solution

R=60(QuantityUnit)  ⟹  f=(R/2)=30(QuantityUnit) ,  h=10(QuantityUnit)  and  s=40(QuantityUnit)

(1/s)+(1/(s'))=(1/f) ⟹ s'= ((((1/f)-(1/s)))^(-1))=120(QuantityUnit)

m=((h')/h)=-((s')/s)  ⟹   h'=-h((s')/s)=-30(QuantityUnit)   and   m=-((s')/s)=-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 h' 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) ,  h=10(QuantityUnit)  and  s=20(QuantityUnit)

(1/s)+(1/(s'))=(1/f) ⟹ s'= ((((1/f)-(1/s)))^(-1))=-60(QuantityUnit)

m=((h')/h)=-((s')/s)  ⟹   h'=-h((s')/s)=+30(QuantityUnit)   and   m=-((s')/s)=+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/s)+(1/(s'))=(1/f)   and   m=((h')/h)=-((s')/s).

It is still true that f=R/2 but now both R and f are negative.  It follows that since s>0 we will always get s'<0 and thus a virtual image.

Example . - Convex Spherical MIrror

Solution

R=-60(QuantityUnit)  ⟹  f=(R/2)=-30(QuantityUnit) ,  h=10(QuantityUnit)  and  s=20(QuantityUnit)

(1/s)+(1/(s'))=(1/f) ⟹  s'= ((((1/f)-(1/s)))^(-1))=-12(QuantityUnit)

m=((h')/h)=-((s')/s)  ⟹   h'=-h((s')/s)=+6(QuantityUnit)   and   m=-((s')/s)=+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/s)+(1/(s'))=(1/f)=0   ⟹  s'=-s

m=((h')/h)=-((s')/s)  ⟹  m=1  and  h'=h

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

. - Images from Refraction

General Sign Conventions in Optics

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

We will now consider cases involving refraction, where the light emerges (ends up) on the opposite side where it originates.  The ray that comes into the optical component (mirror, lens or refracting interface) will be called the incident ray.  The ray leaving the optical component will be called the emergent ray; this is the reflected ray for mirrors or the refracted ray for a refracting interface or lens.

A general sign convention can now be given.  s is positive when the object is on the side of the incident ray, the side where the light originates.   s', f and R are positive when the image, focal point and center, respectively, are on the side of the emergent ray, on the side where the light ends up.

Spherical Interface

The underlying assumption is that all angles are small and all rays hit the interface at a small distance from the central axis, relative to the radius.  When ϕ is a small angle: tan ϕ≃ϕ≃sin ϕ; the ≃ will be replaced with = in this discussion.

sin   (Subscript[θ, 1])=tan   (Subscript[θ, 1])=(h/s)  and  sin  (Subscript[θ, 2])=tan   (Subscript[θ, 2])=((|h'|)/(s'))=-((h')/(s'))

Snell's law relates the heights to the distances and indices.

(Subscript[n, 1]) sin  (Subscript[θ, 1])=(Subscript[n, 2]) sin  (Subscript[θ, 2])  ⟹  (Subscript[n, 1]) (h/s)=(Subscript[n, 2]) ((|h'|)/(s'))  ⟹  ((|h'|)/h)=((Subscript[n, 1])/(Subscript[n, 2]))((s')/s)

We then have a pair of similar triangles, one connecting the object and the center (at R) and the other between the image and center.

((|h'|)/h)=((s'-R)/(s+R))

Combine the two preceding expressions to eliminate |h'|/h, cross multiply and then regroup.

((Subscript[n, 1])/(Subscript[n, 2]))((s')/s)=((s'-R)/(s+R))  
⟹  (Subscript[n, 1])s s'+(Subscript[n, 1])s' R=(Subscript[n, 2])s s'-(Subscript[n, 2])s R  
⟹  (Subscript[n, 1])s' R+(Subscript[n, 2])s R=(Subscript[n, 2])s s'-(Subscript[n, 1])s s'

Dividing by s s'R gives the expression we seek.

((Subscript[n, 1])/s)+((Subscript[n, 2])/(s'))=(((Subscript[n, 2]) - (Subscript[n, 1]))/R)

Since the image is inverted we have |h'|=-h'.   This gives expressions for the image height and magnification.

m=((h')/h)=-(((Subscript[n, 1]) s')/((Subscript[n, 2]) s))

In the diagram s, s' and R are all positive.  The sign convention of R is simple; if the center is on the side of the refracted light, where the index is (Subscript[n, 2]), then it is positive.  If the center is on the side of the incident rays, where the index is (Subscript[n, 1]), then it is negative.  Just as before a positive s' implies a real image and negative means a virtual one.

Example . - A Goldfish in a Bowl

Solution

(Subscript[n, 1])=1.333, (Subscript[n, 2])=1, s=5(QuantityUnit)  and  R=-15(QuantityUnit)

((Subscript[n, 1])/s)+((Subscript[n, 2])/(s'))=(((Subscript[n, 2]) - (Subscript[n, 1]))/R)  ⟹  (1.333/(5(QuantityUnit)))+(1/(s'))=((-0.333)/(-15(QuantityUnit)))  ⟹  s'=((((0.333/(15(QuantityUnit)))-(1.333/(5(QuantityUnit)))))^(-1))=-4.09(QuantityUnit)

Flat Interface

For a flat interface we let R->∞ as we did with mirrors.

((Subscript[n, 1])/s)+((Subscript[n, 2])/(s'))=(((Subscript[n, 2]) - (Subscript[n, 1]))/R)=0  ⟹  ((s')/s)=-((Subscript[n, 2])/(Subscript[n, 1]))

The magnification, then, becomes one.

m=((h')/h)=-((Subscript[n, 1])/(Subscript[n, 2]))((s')/s)=-((Subscript[n, 1])/(Subscript[n, 2]))(-((Subscript[n, 2])/(Subscript[n, 1])))  ⟹  m=((h')/h)=1

Notice from the diagram above that it is clear that m=1.  Also, the diagram shows this can only be a virtual image.

Thin Lens Approximation

A lens is formed from two refracting surfaces.  The light passes from air (n=1) to a medium with index n and then back to air.  Take the two refracting surfaces to have radii (Subscript[R, 1]) and (Subscript[R, 2]).  Using the standard sign convention, take both radii to be positive; this means that their centers are on the side of the light leaving the lens.

Begin with an object to the left of our lens, shown by the green dot; its distance from the first surface is (Subscript[s, 1]).  At the first interface we have

((1)/(Subscript[s, 1]))+(n/(Subscript[s, 1]^′))=((n - 1)/(Subscript[R, 1])).

This gives an image beyond the first surface by (Subscript[s, 1]^′) shown as the black dot.  Note that this image is also beyond the second surface.  When there is a combination of optical elements, in this case refracting surfaces, the image from the first becomes the object for the second.  Since the image from the first surface has yet to form when it hits the second we get what is called a virtual object; this is the odd circumstance when s becomes negative.  (Subscript[s, 2]) is measured from the second interface; its negative value is:

t-(Subscript[s, 2])=(Subscript[s, 1]^′)   ⟹   (Subscript[s, 2])=t-(Subscript[s, 1]^′).

At the second interface we have

(n/(Subscript[s, 2]))+(1/(Subscript[s, 2]^′))=((1 - n)/(Subscript[R, 2])),

where the image, shown with the red dot, is a distance of (Subscript[s, 2]^′) beyond the second surface.

To make this a thin lens we let t->0, giving (Subscript[s, 2])=-(Subscript[s, 1]^′).  Also, take the image and object distances for the lens to be the incoming object distance and the outgoing image distance.

(Subscript[s, 2])=-(Subscript[s, 1]^′), s=(Subscript[s, 1]) and s'=(Subscript[s, 2]^′)

Combining all these expressions above gives an expression relating the object and image positions to the index of the lens' material and the radii of curvature of its surfaces.

(1/s)+(1/(s'))=(n-1) ((1/(Subscript[R, 1]))-(1/(Subscript[R, 2])))

The right-hand side can be identified with 1/f, where f  is defined as the focal length of the lens.  This gives what is called the lensmaker formula:

(1/f)=(n-1) ((1/(Subscript[R, 1]))-(1/(Subscript[R, 2]))).

With this identification the same expressions that applied to both the concave and convex spherical mirrors also apply to the case of a thin lens

(1/s)+(1/(s'))=(1/f)   and   m=((h')/h)=-((s')/s).

Lenses with f>0 are called converging lenses and those with f<0 are called diverging.  Note that converging lenses are always thicker at the center and diverging are thinner at the center.

Example . - The Lensmaker Formula and Eyeglasses

Solution

(1/f)=(n-1) ((1/(Subscript[R, 1]))-(1/(Subscript[R, 2]))).

(a)  n=1.6, (Subscript[R, 1])=30(QuantityUnit)  and  (Subscript[R, 2])=60(QuantityUnit)    ⟹    f=100(QuantityUnit)

(b)  n=1.6, (Subscript[R, 1])=-60(QuantityUnit)  and  (Subscript[R, 2])=60(QuantityUnit)    ⟹    f=-50(QuantityUnit)

Lens Power (in Diopters) =(100/(f (in (QuantityUnit))))

Solution

(a)   f=100(QuantityUnit)   ⟹   +1 diopters

(b)   f=-50(QuantityUnit)   ⟹   -2 diopters

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 s>f  then there is a real image and when the object distance is less than the focal length s<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) ,  h=3.5(QuantityUnit)  and  h'=±210(QuantityUnit).

((h')/h)=-((s')/s)  ⟹ s'=-((h')/h)s

h'=-210(QuantityUnit)   ⟹  s'=-((h')/h)s=60 s

(1/s)+(1/(s'))=(1/f)  ⟹  (1/s)+(1/(60 s))=(1/(12(QuantityUnit)))

12(QuantityUnit) (1+(1/60))=s

s=12.2(QuantityUnit)   and  s'=60 s=732(QuantityUnit)

Example . - Two Lenses and a Virtual Object

Solution

(Subscript[s, 1])=(Subscript[x, 1])=20(QuantityUnit)  and  (Subscript[f, 1])=15(QuantityUnit)   ⟹  (1/(Subscript[s, 1]))+(1/(Subscript[s, 1]^′))=(1/(Subscript[f, 1])) ⟹ (Subscript[s, 1]^′)= ((((1/(Subscript[f, 1]))-(1/(Subscript[s, 1]))))^(-1))=60(QuantityUnit)

(Subscript[x, 0])=(Subscript[x, 1])+(Subscript[s, 1]^′)=20(QuantityUnit)+60(QuantityUnit)=80(QuantityUnit)

(Subscript[s, 2])=(Subscript[x, 1])-(Subscript[x, 0])=50(QuantityUnit)-80(QuantityUnit)=-30(QuantityUnit)

(Subscript[f, 2])=-12(QuantityUnit)   ⟹  (1/(Subscript[s, 2]))+(1/(Subscript[s, 2]^′))=(1/(Subscript[f, 2])) ⟹ (Subscript[s, 2]^′)= ((((1/(Subscript[f, 2]))-(1/(Subscript[s, 2]))))^(-1))=-20(QuantityUnit)

(Subscript[x, f])=(Subscript[x, 2])+(Subscript[s, 2]^′)=50(QuantityUnit)+(-20(QuantityUnit))=30(QuantityUnit)