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Chapter 32

OPTICAL IMAGES

32.1 Mirrors

The point P’ is called the image or the virtual image of P (light does not emanate from it)

The left-right reversal in the mirror is also called the depth inversion (the front and the back are also inverted)

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, , ,i j k i j k× =

i j k× = −

The mirror transforms a right-handed coordinate system

into a left-handed system with

.

θ

θ

The image of an arrow. To locate the image of any point P, one can follow two rays:

(1)perpendicular to the mirror (is reflectedonto itself)

(2) ray at some angle with the normal

to the mirror (is reflected at an equal angle

).The intersection of these two rays gives the image P’. The image is at the same distance behind the mirror as the origin in front, is upright, and is the same size as the origin.

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A set of mirrors provides multiple images. P1’ is the image of P in mirror 1, P2’ is the image of P inmirror 2, and P1,2’ is the image of P with reflections first in mirror 1 and then in mirror 2.

A ray reflected from two perpendicular mirrors is reflected back along a parallel path irrespective of an angle of incidence

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Spherical mirrors

The rays from a point source P on the axis of concave mirror converge in the (real) image P’ (the light actually emanate fromthis point!) and then continue as if it is a real point source

Spherical aberration. Only the rays that are close to the axis (paraxial rays) converge in the image point; the faraway rays blur the image. To sharpen theimage one must block non-central parts of the mirror.

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Finding location of the image. The ray through the center of the sphereis reflected onto itself. The ray striking at point A at angle with θwith the normal is reflected at angle θwith the normal. The geometry dictates:

1 1 2'

, 2 2 ,/ , / , / ' s s rl s l r l s

β α θ γ α θ β α γα β γ= + = + ⇒ = +≈ ≈ ≈ + =⇒

This equation relates the distance to the image to the distance to the object and the radius of the sphere

Solving this equation for s’ yields

2 2 /' sr rs r r ss − −= ≡

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Distance of the image from the axis.Two right triangles in the Figure aresimilar. Thus

( )'/ '/ ' '/y y s y ys s s= − ⇒ = −

(the negative sign means that P and P’ are on the opposite sides of the axis)

' / 2s r≈( )' '/ 0y y s s= − →

When the object is far away from the mirror, s >> r, 1/s << 2/r and the image is at the point

which is called the focal point F of the mirror ( ).

Using the distance to the focal point f = r/2 instead of r, one gets the mirror equation

1/ 1/ ' 1/s s f+ =

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Parallel rays striking a concave mirror are reflected to a point on the focal plane. In reverse, if a source is on the focal plane the reflected rays are parallel to each other. This illustrates the principle of reversibility of reflection (and refraction)

Example 1. An object is 24 cm from concave mirror of radius 16 cm, 2 cm above the axis. Find the position of the image and the focal length of the mirror.

The focal length of the mirror f = r/2 = 8 cm. Using the mirror equation, one gets:

( ) ( )

1 11/ ' 1/ 1/ (1/8 1/ 24) 0.0833' 12 ; ' '/ 2 12 / 24 1s f s cm cm

s cm y y s s cm cm

− −= − = − =

= = − = − ⋅ = −

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Ray diagrams for mirrors

Three principal rays:

(1) Ray parallel to the axis after reflection goes through the focal and image point

(2) Ray going through the focal point, after reflection goes parallel to the axis through the image point

(3) Ray going through the center C is normal to the mirror and, after reflection, reverses itself and goes through the image point.

The image point lies at the intersection of these three rays. Any two of them can be used in construction for finding the image point.

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Typical configurations

Central point is between the object and the mirror: The image is between the center and the focal point, is reversed, and issmaller than the object

Object is between the central point and the mirror: The image is behind the mirror, is direct (not reversed), and is bigger than the object (here s’ is negative)

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Sign conventions for the mirror equation

(1) s is positive if the object is on the incident-light side of the mirror

(2) s’ is positive if the object is on the reflected-light side of the mirror

(3) r (or f) is positive if the mirror is concave (center of curvature is on the reflected-light side of the mirror)

The lateral magnification m is the ratio of the image size to the object size:

'/ '/m y y s s= ≡ −(m is negative if the image is inverted).

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Convex mirrorsare described by the same mirror equation with the proper sign conventions

Ray diagram for a convex mirror. In onstruction, one should use the same (three) principal rays

Example 2. An object 1.5 cm high is 24 cm away from a convex mirror of radius 12 cm. Find the image.

The focal distance is –12/2 cm = - 6 cm. The mirror equation yields:

( )( ) ( )

( )

1 11/ ' 1/ 1/ 1/ 6 1/ 24 0.2083 ,

' 4.8 ; ' '/ 1.5 4.8 / 24 0.3 ,

'/ 4.8 / 24 0.2

s f s cm cm

s cm y y s s cm cm

m s s

− −= − = − − = −

= − = − = − ⋅ − =

= − = − − =

The image is behind the mirror (virtual image; s’ < 0), is upright (y’, m > 0), and is smaller than the object (|m| < 1).

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32.2 LENSESImages formed by refraction

The ray construction for cylinder with spherical surface made of material with refraction index n2.Light comes from the medium with refraction index n1. The angles are related by Snell’s law. For paraxial refraction, the angles are small. The distance to the image is

1 2 2 1'

n n n ns s r

−+ =In refraction, the real images are formed in back of the surface (refracted-light side), the virtual images – in front of the surface.

(1) s is positive for objects on the incident side(2) s’ is positive for images on refracted-light side(3) r is positive if the center is on the refracted-light side

Rules

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Example 3. Derive an expression for magnification m = y’/y by a spherical refracting surface.

Snell’s law in a small angle approximation has a form

1 1 2 2 1 1 2 2sin sinn n n nθ θ θ θ= ⇒ =

From geometry, '

1 1 2 2 'tan ; tany ys sθ θ θ θ≈ = ≈ = −

Combining,

1

1

2

2' '' ss

s nyy snm θ

θ− = −==

Example 4. An object is 3 m below the flat surface of water. Where is the image?

In equation 1 2 2 1'

n n n ns s r

−+ = n1 = 1.33 (water), n2 = 1 (air), s = 3 m, r is infinite(flat surface). Thus, 1 1.331

' ' , ' /1.33 2.5ns s s s s m≡ = − = − = −

The negative sign means that the image is virtual – on the opposite side from the refracted-light side (under water). The magnification is

1

2

' 1.33 2.53 1n s

n sm ⋅= − = = which is not surprising (flat surface!)

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Example 5. The object is in the air, 20 cm from a spherical container with water (r = 12 cm). Where is the image and what is the magnification?

The main equation is again

1 2 2 1'

n n n ns s r

−+ =n1 = 1.0 (air), n2 = 1.33 (water), s = 0.2 m, r = 0.12 m. Then

( ) 1 11.33 1.33 1.0 1.0' 0.12 0.2

1.332.25

2.25 ,

' 0.59s m ms m m

− −−= − = −

= − = −

meaning that the image is in the air, in front of the sphere.The magnification is,

1

2

'' 0.590.2 1.33 2.22s ny

y snm = = − = =i

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Thin Lenses

A lens has two curved surfaces. The refraction at each surface should be considered separately.

Example 6. Consider a thin lens of index of refraction n in the air. Thecurvatures are r1 and r2, the object is at distance s from the (first surfaceof the) lens. Find the image.

The distance s1’ of the image due to the refraction at the first surface is

' 11

11 n ns rs

−+ =The first surface image becomes the object for the second surface. This light is refracted at the second surface with the “object” being, in this case, at the negative distance s1’ from it (“virtual object” to the left of the second surface):

' 21

11'

n ns rs

−−+ =

Adding these equations eliminates s1’,

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( )1 1 22

1 1 11'

1 1 1 11,s s f f r rn nr r n− −+ ≡ + = = − −

This is the thin-lens equation.

The equation for the focal length f in terms of the radii of the surfaces is called the lens-maker’s equation. The focal length of a thin lens gives the image distance when the object distance is infinite.

For a lens in this figure, r1 is positive and r2 is negative, so f is positive (a double convex orconverging lens or positive lens).In general, the image distance s’ is positive when the image is on the refracted-light side of the lens

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This figure shows a double concave or diverging lens (a negative lens) the plane wave diverges after refraction. The focal distance is negative. Any lens that is thinner in the middle is negative.

Example 7. Find the focal length of this double-convex lens.

( ) ( )1 2

1 1 1 1 1 1 110 15 121 1.5 1f r r cm cmn −

= − − = − − =

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Each lens has two focal points: convergence (or divergence) points for light coming fromthe right or left. Both points are at the same distance, but on the opposite sides of the lens defined as the first and second focal points depending on direction of the incident light. For a positive lens, the first focal point is on the incident-light side.

If parallel rays are incident at a small angle with the axis, the light is focused at a point in the focal plane a distance ffrom the lens.

The reciprocal of the focal length is called the power of lens, P = 1/f. When f is in m, the power is measured in reciprocal meters, m-1, which are called diopters [D]

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Example 8. Find the focal length and the power of this lens.

Here both are r1 and r2 are positive:

( ) ( )1 2

1 1 1 1 10.1 0.131 1.5 1

1.15 ; 1/ 0.87 87f r rP n

D f P m cm

= = − − = − − = = = =

In the lab, it easier to measure the focal length f than measure r1 and r2 and then calculate f.

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Ray diagramIn construction, it is easier to follow three principal rays which converge to the image point:

(1) The central ray (undeflected)(2) The focal ray – emerges parallel to the axis(3) The parallel ray – the emerging ray is directed toward (or away from) the second focal point.

The ray diagram for a negative lens is here:

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The lateral magnification of the lens is

'/ '/m y y s s= = −Negative magnification means that the image is inverted.

Example 9. An object h = 0.8 cm high is placed 6 cm from a double convex lens with f = 8 cm. Find the image graphically and algebraically, find magnification, and determine whether the image is real or virtual.

The object is between the focus and the lens. The diagram looks like this.

11 1 1 1 1 16 ' 8 ' 8 6, 0.042 ,

' 24 ; '/ 4cm s cm s cm cm cm

s cm m s s

−+ = = − = −

= − = − =

The height of the image h’ = mh = 4 x 0.8 cm = 3.2 cm

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Combination of lensesExample 10. The object is 4 cm from a double convex lens with f = 12 cm. The second lens with f = 6 cm is placed 12 cm to the right of the first. Find the final image.

By accident (or by design?) the second lens is placed into the focus of the first Thus, the parallel ray of the first lensis the central ray for the second. The focal ray of the first lens is always the parallel ray of the second. The construction is simple.

The object for the second lens is the (virtual) image of the first located at1 1 1 1 1 1 14 ' 12 ' 12 4 6, , ' 6 ; '/ 1.5cm s cm s cm cm cm s cm m s s+ = = − = − = − = − =

This image / object is s2 = 6 cm + 12 cm = 18 cm away from the second lens,

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1 1 1 1 1 1 12 2 2 218 6 ' 6 18 9, , ' 9 ; '/ 0.5cm cm s cm cm cmss cm m s s+ = = − = = = − = −

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Example 11. Two lenses of focal length 15 cm each are 22.5 cm apart. Find the image of an object 22.5 cm from one of the lenses.

The first construction yields the distance to the “first image”:

1 1 122.5 ' 15

1 1 1 1' 15 22.5 45

,

,

' 45 ;'/ 2

cm s cm

s cm cm cm

s cmm s s

+ =

= − =

== − = −

The rays for the second construction are continuation of the first. The final image is at

( ) 2 2

1 1 1 1 1 1 1' 15 ' 15 22.5 945 22.5

2 2 2 2 1 2

, ,

' 9 ; '/ 0.4; 0.8s cm s cm cm cmcm

s cm m s s m m m− + + = = + =

= = − = = ⋅ = −

The final image is inverted and real (see the diagram).

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Compound lenses

When two thin lenses are compounded (brought close together), the effective focus length is

1 2 1 21/ 1/ 1/ ,eff efff f f P P P= + = +

Proof. For the first lens,

1

1 1 1' .s s f+ =

For the second one, the object distance is the negative of the first imagedistance,

2 2

1 1 1' .s s f− + =

Adding these two equations, we get

( ) ( ) ( )2 2 1 2

1 1 11 11 1 1' '

1 .effs s s s s f f fs− + = + =+ + + =

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Aberrations

Blurring that results fromrays not focused at a single focus point is calledaberration. The aberrationis usually caused by largevalues of angles between the rays and the axis. The circle of least confusion is at point C where the diameter of the image of a point object is minimum. This type of aberration in a thin lens is called spherical aberration.

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32.4 Optical InstrumentsThe Eye

The light is focused by the cornea, with assistance from the lens, on the retina. When the object is far away, system focal length is maximum, about 2.5 cm (distance from the cornea to retina). For closer objects, thecurvature of the lens increase thus decreasing the focal length so that image is still focused on the retina – accommodation. In the lens equation for the eye

1 1 1's s f+ =

the distance to the image s’ is fixed and changing s requires changing f. The closest point for which the lens can still focus the image is called the near point. This distance changes with age and from person to person; the “standard” value is defined as 25 cm.

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Farsightedness Nearsightedness

' '/y ys s= −The size of the image is

Since s’ = 2.5 cm and does not change, the bigger the distance to the object s, the smaller the image.

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Example 12. The near-point distance of a person is 75 cm. The reading glasses, that are very close to the eye (a compound lens) correct it to 25 cm. That is, the object put 25 cm in front of eye glasses forms a virtual image 75 cm from the eye. Describe the glasses.

The lens equation for the glasses is

1 1 1 1 1 2' 25 75 75 ; 37.5 0.375 ;

2.67 ; 75 / 25 3s s f cm cm cm f cm m

P D m−+ = = + = = =

= = − − =Such glasses not only change the near point, but also magnify the object (the fact well-known to everyone who has them)!

A simple magnifier – a converging lens put close to an eye if the object is closer to the lens than its focal length.

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The (compound) microscopeis used to look at very small objects at small distances.The lens close to the object –objective – forms a real image. The eyepiece is a simplemagnifier. The eyepiece is placed so that the image formed by the objective is in its first focal plane (viewing the image at infinity). The distance between the second focus of the objective and the first focus of the eyepiece is the tube length(about 16 cm).

The magnification of the objective is tan / '/ '/ /o o oy f y L m y y L fβ = = − ⇒ = = −

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The magnification of the eyepiece (simple magnifier) is

/ , / ,

/ /o np e

e o np e

y x y fM x fθ θ

θ θ

= =

= =

where xnp is the near point distance.

The overall magnification is

0 /e np o eM m M Lx f f= = −

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Example 13. The focal lengths of an objective and an eyepiece are 1.0 and 1.8 cm. The near point of the viewer is 25 cm. Find the magnifying power and the position of the object if the final image is viewed at infinity. The distance between the lenses is 22 cm.

The tube length is the separation between the lenses minus the focaldistances,L = (22 – 1.0 - 1.8) cm = 19.2 cm

The magnification is

( )0 /

(19.2 25) / 1.0 1.8 267e np o eM m M Lx f f= = −

= − ⋅ ⋅ = −For the final image to be viewed at infinity, the image for the objective should be in the focal plane of the eyepiece. Then, from the figure, the image distance for the objective is 0' 1.0 19.2 20.2s f L cm cm cm= + = + =

The lens equation for the objective 1/ 1/ ' 1/ os s f+ = gives the position of theobject:

( ) ( ) 11/ 1/ 1.0 1/ 20.2 0.95 ; 1.05s cm cm cm s cm−= − = =

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Telescopeis used to view large objects very far away by creating a smaller, but much closer real image of the object.

o ef f+/e oM θ θ=

The purpose of the objective is not to magnify, but to produce the image that is close to us. This image is in the focal plane of the eyepiece so that it is viewed at infinity. Since this image is in the second focal plane of the objective (the object is at infinity!), the separation between the lenses is The magnifying power is the angular magnification According to the figure,

' '0

0

tan ; tan ;

/ /o e

y y yo e es f f

o eeM f f

θ θ θ θ

θ θ

−= = ≈ =

−==

≈

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Example 14. The distance between the lenses is 40 cm. The focal length of the eyepiece is 4 cm. Find the magnifying power.

40 4 36of cm cm cm= − =

/ 36 / 4 9o eM f f= − = − =

The distance between the lenses is the sum of focal lengths. Therefore,

and the magnifying power is

.

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Review of Chapter 32

An image is real if actual light rays converge to the image point (in front of the mirror or on refracted-light side of a refractor). An image is virtual if only extensions of the actual rays converge to it.

An object is real if actual light rays diverge from it (actual object or a real image). An object is virtual if only extensions of the actual rays diverge from it.

Mirror equation

1/ 1/ ' 1/ ; / 2s s f f r+ = =where the focal length is the image distance when the object is at infinity

Lateral magnification '/ '/M y y s s= = −

Images can be easily located using the ray diagram for three principal rays: the parallel ray, the central (radial) ray, and the focal ray.

Sign conventions: s is positive if the object is on the incident side of the mirror; s’ is positive if the image is on the reflected-light side; r and f are positive if the mirror is concave (the center is on the reflected-light side).

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Refraction. At a single surface

1 2 2 11 2' , '/ '/n n n n

s s r m y y n s n s−+ = = = −Sign conventions: s is positive if the object is on the incident side; s’ is positive if the image is on the refracted-light side; r is positive if the center is on the refracted-light side.

Thin lenses. Thin lens and lens-maker’s equations

( )1 2 1 2

1 11 1 1 1 1 1' , 1n n

s s r r f f r rn− − + = + ≡ = − − Incident rays parallel to the axis emerge directed towards or away from the first focal point; incident rays directed towards or away from the second focal point emerge parallel with the axis.

Power and magnification 1/ ; '/ '/P f m y y s s= = = −Images can be easily located using the ray diagram for two of three principal rays: the parallel ray, the central (radial) ray, and the focal ray.

Sign conventions: s > 0 if the object is on the incident side; s’ > 0 if the image is on the refracted-light side; f > 0 for converging lens, f < 0 for diverging lens.

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The Eye. Light is should be focused on the retina, about 2.5 cm from the cornea. F for the lens changes depending on the distance from the object; the closest distance for which the focusing is still possible – the near point, about 25 cm. The closer the object, the larger is the image on the retina.

The simple magnifier. A lens with a focus length that is smaller than the near point, / /e o npM x fθ θ= =The compound microscope has two lenses – the objective and the eyepiece. The object is outside the focal point of the objective with an image in the focal plane of the eyepiece. The eyepiece is a simple magnifier.The magnifying power 0 /e np o eM m M Lx f f= = −where the tube length is the distance between the second focal point of the objective and the first of the eyepieces.

The telescope also has two lenses which are separated by a distance equal to sum of their focal lengths. The objective creates a smaller image than the object but at a much closer distance. The eyepiece is a simple magnifier.The magnifying power 0/ /e o eM f fθ θ= = −