Nature of Light

Ancient times & Middle Ages: light consist of stream of particles corpuscles 19th century: light is a wave (EM wave) Geometrical 20th century: several effects associated with the emission and absorption of light revel that it also has a particle aspect and that the energy carried by light waves is packaged in Chapter 24 discrete bundles called photons. Quantum electrodynamics – a comprehensive theory that includes both wave and particle properties. The propagation of light is best described by a wave model The emission and absorption by atoms and nuclei requires a particle approach

Wave Fronts Rays

Wave Front – a convenient concept to describe wave It is very convenient to represent propagation a light wave by rays rather than A wave front is the locus of all adjacent points at which by wave fronts the phase of vibration of the wave is the same A ray is an imaginary line along the direction of travel of the wave

The branch of optics for which the ray description is adequate is called geometric optics The branch dealing specifically with wave behavior is called physical optics spherical and planar wave fronts and rays

Reflection and Refraction

When a light ray travels from one medium to another, Part 1 part of the incident light is reflected and part of the light is transmitted at the boundary between the two media. The transmitted part is said to be refracted in the The of light second medium.

incident ray reflected ray

refracted ray

1 Types of Reflection

If the surface from which the light is reflected is smooth, then the light undergoes (parallel rays will all be reflected in the same directions). If, on the other hand, the surface is rough, then the light will undergo diffuse reflection (parallel rays will be reflected in a variety of directions) Eduard Manet – A Bar at the Foleis-Bergère (1882)

Types of Images for and Lenses The Law of Reflection A real image is one in which light actually passes through the image point For specular reflection the incident angle θi equals the reflected angle θ : Real images can be displayed on screens r A virtual image is one in which the light does not pass through the image point θi = θr The light appears to diverge from that point Virtual images cannot be displayed on screens

The angles are measured relative to the normal as shown.

Plane Mirrors Forming Images with a Plane

A mirror is a surface that can reflect a beam of light in A is simply a flat mirror. one direction instead of either scattering it widely in Consider an object placed at point P in front of a plane many directions or absorbing it. mirror. An image will be formed at point P´ behind the mirror.

do = distance from object to mirror

di = distance from image to mirror

ho = height of object

hi = height of image ho hi For a plane mirror: do di

do = di and ho = hi

2 Plane Mirrors What is wrong with the painting?

A plane mirror image has the following properties: 9 The image distance equals the object distance 9 The image is unmagnified 9 The image is virtual 9 The image is not inverted

Conceptual Checkpoint

To save expenses, you would like to buy the shortest mirror that will allow you to see your entire body. Should the mirror be equal to your height? Part 2 Does the answer depend on how far away from the mirror you stand? Spherical Mirrors

Spherical Mirrors Concave Mirror, Notation

A spherical mirror is a mirror concave whose surface shape is • The mirror has a radius spherical with radius of curvature of curvature of R • Its center of curvature is R. There are two types of the point C spherical mirrors: concave and • Point V is the center of convex. the spherical segment We will orient the mirrors so that • A line drawn from C to V is called the principle the reflecting surface is on the axis of the mirror left side of the mirror. The object will be to left of the mirror itself. convex

3 Focal Point Focal Length Shown by Parallel Rays

When parallel rays are incident upon a spherical mirror, the reflected rays intersect at the focal point F. For a concave mirror, the focal point is in front of the mirror. For a convex mirror, the focal point is behind the mirror. The incident rays diverge from the convex mirror, but they trace back to the focal point F.

Focal Length Ray Tracing

The focal length f is the distance from the surface of the We will use three mirror to the focal point. It can be shown that the focal principal rays to length is half the radius of curvature of the mirror. determine where an image will be Sign Convention: the focal length is negative if the focal located. point is behind the mirror.

For a concave mirror, f = ½R The parallel ray (P ray) reflects For a convex mirror, f = −½R (R is always positive) through the focal point. The focal ray (F ray) reflects parallel to the axis, and the center-of-curvature ray (C ray) reflects back along its incoming path.

Ray Tracing – Examples More examples: Images from spherical mirrors

concave convex

Real images form on the side of the mirror where the object is, and virtual images form on the Virtual image Real imageopposite side

4 Ray Diagram for Concave Mirror, p > R

• The image is real • The image is inverted • The image is smaller than the object

Ray Diagram for a Concave Mirror, p < f Ray Diagram for a Convex Mirror

• The image is virtual • The image is virtual • The image is upright • The image is upright • The image is larger than the object • The image is smaller than the object

Spherical Aberration Image Formed by a Concave Mirror

• Mirror is not exactly spherical • Geometry shows the • This results in a blurred relationship between image the image and object • This effect is called distances spherical aberration 112 + = pq R

This is called the mirror equation

5 problem The Mirror Equation Sign Conventions: di is positive if the image is in Example 1 The ray tracing technique front of the mirror (real image) shows qualitatively where the di is negative if the image is image will be located. The behind the mirror (virtual An object is placed 30 cm in front of a concave mirror of distance from the mirror to the image) radius 10 cm. Where is the image located? Is it real or image, di, can be found from virtual? Is it upright or inverted? What is the magnification f is positive for concave mirrors the mirror equation: of the image? f is negative for convex mirrors 1 1 1 + = m is positive for upright images 11111 1 += += d 6 do di f m is negative for inverted images m =−i =− =−0.2 dd0 ii f30 d⎛⎞10 ⎜⎟+ d0 30 2 image magnification. If m is ⎝⎠ Image is smaller, real and inverted. do = distance from object to negative, the image is inverted 111 530× mirror =−or di = =6 cm (upside down). di 530 305− di = distance from image to mirror image height h d m = = i = − i f = focal length object height ho do

problem problem Example 2 Example 3

An object is placed 5 cm in front of a convex mirror of An object is placed 3 cm in front of a concave mirror of radius focal length 10 cm. Where is the image located? Is it real 20 cm. Where is the image located? Is it real or virtual? Is it or virtual? Is it upright or inverted? What is the upright or inverted? What is the magnification of the image? magnification of the image?

11111 1 d 3.33 + =+=− m =−i = =0.67 11111 1 d 4.3 dd0 ii f510 d += += i d0 5 20 m =− = =1.43 dd0 ii f3 d⎛⎞ d 3 111 ⎜⎟ 0 or Image is smaller, virtual, and upright. ⎝⎠2 =− − Image is larger, virtual, and upright. di 10 5 111 310× = −or d =− =−4.3 cm 510× d 10 3i 7 d =− =−3.33 cm i i 15

problem Problem A concave mirror produces a virtual image that is three times as tall as the object. (a) If the object is 22 cm in front of the mirror, what is the image distance? (b) What Part 3 is the focal length of this mirror? The Refraction of Light d i 111111 am.) =− b.) += −= d0 dd0 i f22 66 f d 366=−i d =− cm66× 22 22 i f ==33 cm 66− 22

6 The Refraction of Light Snell’s Law In general, when light enters a new material its direction will change. The angle of refraction θ2 is related to the angle of The speed of light is different in different materials. We incidence θ by Snell’s Law: 1 sinθ sinθ define the index of refraction, n, of a material to be the ratio 1 = 2 of the speed of light in vacuum to the speed of light in the v1 v2 where v is the velocity of light in the medium. material: c n = Normal line v θ When light travels from one medium to another its velocity Snell’s Law can also be written as 1 and wavelength change, but its frequency remains n1 sinθ1 = n2 sinθ2 Air constant.

For a vacuum, n = 1 The angles θ1 and θ2 are Glass measured relative to the line For other media, n > 1 θ normal to the surface between 2 n is a unitless ratio the two materials.

Application – Day and Night Settings on Auto Example: Which way will the rays bend? Mirrors n = 1.4 n = 2

n = 1.6 n = 1.2

• With the daytime setting, the bright beam of reflected light is directed into the driver’s eyes Which of these rays can be the refracted ray? • With the nighttime setting, the dim beam of reflected light is directed into the driver’s eyes, while the bright n1 sinθ1 = n2 sinθ2 beam goes elsewhere

problem A Common Mirage Problem

You have a semicircular disk of glass with an index of refraction of n = 1.52. Find the incident angle θ for which the beam of light in the figure will hit the indicated point on the screen.

7 problem

nn1122sinθ = sinθ Total Internal Reflection 5 1×=× sinθθθ122 1.52 sin tan = When light travels from a medium with n > n , there is an 20 1 2 angle, called the critical angle θc, at which all the light is o Therefore θ2 = 14 reflected and none is transmitted. This process is known as o total internal reflection. sinθ1 =× 1.52 sin14

o Therefore θ1 = 21.6

The incident ray is both reflected and Total Internal Reflection refracted.

Critical Angle Fiber Optics

A particular angle of • An application of internal incidence will result in reflection • Plastic or glass rods are an angle of refraction of used to “pipe” light from 90° one place to another – This angle of • Applications include incidence is called the – medical use of fiber optic critical angle cables for diagnosis and correction of medical problems – Telecommunications n2 sinθc = for n1 > n2 n1

problem problem Problem At the first reflection, o nmed sin 45= 1 A ray of light enters the long side of a 45°-90°-45° prism This is the same for the second reflection. and undergoes two total internal reflections, as indicated Therefore, in the figure. The result is a reversal in the ray’s 1 direction of propagation. Find the minimum value of the sin 45o ≥ or n prism’s index of refraction, n, for these internal med 1 reflections to be total. n ≥=2 med sin 45o

nmed = 1.414

8 ? ? Question Question

Sometimes when looking at a window, one sees two reflected A student claims that, because of atmospheric refraction, the images, slightly displaced from each other. sun can be seen after it has set and that the day is therefore What causes this effect? longer than it would be if the earth had no atmosphere.

What does the student mean by saying the sun can be seen after it has set?

Does the same effect also occur at sunrise?

Lenses Part 4

Lenses Light is reflected from a mirror. Light is refracted through a lens.

Focal Point Ray Tracing for Lenses

The focal point of a lens is the place where parallel rays Just as for mirrors we use incident upon the lens converge. three “easy” rays to find the image from a lens. The lens is assumed to be thin.

The P ray propagates parallel to the principal axis until it encounters the lens, where it is refracted to pass through the focal point on the far side of the lens. The F ray passes through diverging lens the focal point on the near side of the lens, then leaves the converging lens lens parallel to the principal axis. The M ray passes through the middle of the lens with no deflection.

9 The focal length for a thin lens Ray Tracing Examples

1 ⎛ 1 1 ⎞ = (n −1)⎜ − ⎟ f ⎝ R1 R2 ⎠

problem The Thin Lens Equation Example 4 The ray tracing technique shows qualitatively 1 1 1 An object is placed 20 cm in front of a converging lens of where the image from a lens will be located. + = focal length 10 cm. Where is the image? Is it upright or The distance from the lens to the image, di, d d f can be found from the thin-lens equation: o i inverted? Real or virtual? What is the magnification of the image? Sign Conventions: 111 111 111 += += =− d is positive for real images (on the opposite side of the lens i dd0 iii f20 d 10 d 10 20 from the object) 10× 20 d ==20 cm di is negative for virtual images (same side as object) i 20− 10 f is positive for converging (convex) lenses Since di is positive, the image is real. f is negative for diverging (concave) lenses di m is positive for upright images m =− =−1 Therefore the image in inverted . d0 m is negative for inverted images

problem problem Example 5 Example 6

An object is placed 5 cm in front of a converging lens of An object is placed 8 cm in front of a diverging lens of focal length 10 cm. Where is the image? Is it upright or focal length 4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of inverted? Real or virtual? What is the magnification of the image? the image? 111 111 111 11111111111⎛⎞ += += =− += +=− =−−=−+⎜⎟ dd0 iii f84 d d 4848⎝⎠ dd0 iii f510105 d d ⎛⎞48× 10× 5 di =−⎜⎟ = −2.67 cm d ==−10 cm ⎝⎠48+ i 510− Since d is negative, the image is virtual. Since d is negative, the image is virtual. i i d 2.67 m =−i = =0.33 Therefore the image in upright and smaller . di 10 d 8 m =− = =2 Therefore the image in upright . 0 d0 5

10 problem problem Problem

(a) Determine the distance from lens 1 to the final image for the system shown in the figure. (b) What is the magnification of this image?

111 11 1 1 11⎛⎞ 11 First lens += +=− =−−=−+⎜⎟ dd0 iii f24 d 7 d 7 24⎝⎠ 7 24 ⎛⎞724× When you have two di =−⎜⎟ = −5.42 cm lenses, the image of ⎝⎠724+ 111 1 11 11 1 Second lens += += =− the first lens is the dd f()35+ 5.42 d 14 d 14 40.42 object for the 0 iii ⎛⎞14× 40.42 di ==⎜⎟21.42 cm second lens. ⎝⎠40.42− 14 Distance from lens 1 is 21.42+= 35 56.42 cm

⎛⎞⎛⎞ddii⎛⎞⎛⎞5.42 21.42 mmm=×12 =−⎜⎟⎜⎟ ×− =⎜⎟⎜⎟ ×− =−0.12 dd24 40.42 ⎝⎠⎝⎠0012⎝⎠⎝⎠

Dispersion

In a material, the velocity of light (and therefore the index of refraction) can depend on the wavelength. This is Part 5 known as dispersion. Blue light travels slower in glass and water than does red light.

Dispersion As a result of dispersion, different colors entering a material will be refracted into different angles. Dispersive materials can be used to separate a light beam into its spectrum (the colors that make up the light beam). Example: prism

The Rainbow Observing the Rainbow

• If a raindrop high in the sky is observed, the red ray is seen • A drop lower in the sky would direct violet light to the observer • The other colors of the spectra lie in between the red and the violet

11 Using Spectra to Identify Gases Spherical Aberration

All hot, low pressure gases emit their own characteristic spectra • Results from the focal The particular wavelengths emitted by a gas serve as points of light rays far “fingerprints” of that gas from the principle axis are different from the focal points of rays Some uses of spectral analysis passing near the axis Identification of molecules • For a mirror, parabolic Identification of elements in distant stars shapes can be used to correct for spherical Identification of minerals aberration

problem Chromatic Aberration Problem The index of refraction for red light in a certain liquid is 1.320; the index of refraction for violet light in the same liquid is 1.332. • Different wavelengths of Find the dispersion (θv – θr) for red and violet light when both are light refracted by a lens incident on the flat surface of the liquid at an angle of 45.00° to focus at different points the normal. – Violet rays are refracted more than red rays SNELL'S LAW – The focal length for red light is greater than the focal length nnsin1122rrrsinθθ= for violet light n sinθ sin 45o • Chromatic aberration can be 11r o sinθθ22rr== =32.39 and minimized by the use of a n2r 1.32 combination of converging n sinθ sin 45o and diverging lenses 11b o sinθθ22bb== =32.06 n2b 1.332

ooo θθ22rb−=32.39 − 32.06 = 0.33

The Human Eye

Part 6

Instruments

12 Simple Magnifying Lens Compound Microscope

Mirrors Lenses Convex Mirror Concave Lens Object Image Image Image Object Image Image Image location orientation size type location orientation size type Arbitrary Upright Reduced Virtual Arbitrary Upright Reduced Virtual

Concave Mirror Convex Lens Object Image Image Image Object Image Image Image location orientation size type location orientation size type Beyond C Inverted Reduced Real Beyond F Inverted Reduced or enlarg. Real C Inverted Same as object Real Just beyond F Inverted Approaching Infinity Real Between F & C Inverted Enlarged Real Just inside F Upright Approaching Infinity Virtual Just beyond F Inverted Approaching Infinity Real Between F & lens Upright Enlarged Virtual Just inside F Upright Approaching Infinity Virtual Between F & mirror Upright Enlarged Virtual

You look downward at a coin that ? lies at the bottom of a pool of liquid Question with depth d and index of A concave mirror (sometimes surrounded by light) is often refraction n. used as an aid for applying cosmetics to the face. Because you view with two eyes, which intercept different rays of Why is such a mirror always concave rather than convex? light from the coin, you perceive What considerations determine its radius of curvature? the coin to be where extensions of the intercepted rays cross, at depth da instead of d. Assuming that the intercepted rays are close to a vertical axis through the coin, show that da = d/n. (Hint: Use the small-angle approximation that sinθ= tanθ = θ.)

13 ? problem Question Figure shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 1.0 cm along the horizontal A person looks at her reflection in the concave side of a direction, but the vertical direction is not to the same scale. shiny spoon. (a) Using only the ray shown, decide what type of lens this is. Is this image right side up or inverted? (b) What is the focal length of the lens? (c) Locate the image by drawing the other two principal rays, What does she see if she looks in the convex side? (d) Calculate where the image should be, and compare with solution in (c).

problem

Figure shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 1.0 cm along the horizontal direction, but the vertical direction is not to the same scale. (a) Using only the ray shown, decide what type of lens this is. (b) What is the focal length of the lens? (c) Locate the image by drawing the other two principal rays, (d) Calculate where the image should be, and compare with solution in (c).

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