chapter 17. geometric optics the study of how light behaves as it interacts with different media...

Chapter 17

Geometric Optics The study of how light behaves as it interacts with different media Assumes straight-line propagation Considers light as individual rays (one at a time)

Optical Devices Lenses bend (refract) light rays Converging focus rays and enlarges an image (CONVEX) Diverging spread light rays and reduces an image (CONCAVE)

Optical Devices Mirrors reflect rays and can be curved to redirect light Prisms reflect/refract light, sometimes to varying degrees (also known as Dispersion).

Reflection of Light A ray of light, the incident ray, travels in a medium When it encounters a boundary with a second medium (a discontinuity), some of the incident ray is reflected back into the first medium, becoming a reflected ray. This means some of it is directed backward into the first medium

Law of Reflection The normal is a line perpendicular to the surface It is at the point where the incident ray strikes the surface The incident ray makes an angle of 1 with the normal ( i ) The reflected ray makes an angle of 1 ' with the normal ( r) The angle of incidence is equal to the angle of reflection first part of the Law of Reflection i = r

Law of Reflection, cont A ray is a line drawn in space corresponding to the direction of flow of radiant energy All angles are measured from the perpendicular (or normal) to the surface The second part of the Law of Reflection the incident ray, the perpendicular to the surface, and the reflected ray all lie in a plane called the plane-of-incidence.

Wave Front / Ray A wave front is a surface passing through points of a wave that have the same phase and amplitude The rays, corresponding to the direction of the wave motion, are perpendicular to the wave fronts

Specular Reflection Specular reflection is reflection from a smooth surface The reflected rays are parallel to each other

Diffuse Reflection Diffuse reflection is reflection from a rough surface The reflected rays travel in a variety of directions Diffuse reflection makes the dry road easy to see at night

Internal and External Reflection When light propagates through a medium encounters a change in medium (discontinuity) some light scatters in a backward direction (reflection) Glass Air Beam IIBeam I Light going from less to more dense medium external reflection Light going from more to less dense medium internal reflection

Refraction of Light When a ray of light traveling through a transparent medium encounters a boundary leading into another transparent medium, part of the ray is reflected and part of the ray enters the second medium The ray that enters the second medium is bent at the boundary This bending of the ray is called refraction

Refraction of Light The incident ray, the reflected ray, the refracted ray, and the normal all lie on the same plane The angle of refraction, 2, depends on the properties of the medium

Following the Reflected and Refracted Rays Ray is the incident ray Ray is the reflected ray Ray is refracted into the lucite Ray is internally reflected in the lucite Ray is refracted as it enters the air from the lucite

The Index of Refraction When light passes from one medium to another, it is refracted because the speed of light is different in the two media The ratio of the speed of an electromagnetic wave in vacuum to that in matter is the index of refraction n: n = c/ v For a vacuum, n = 1 For other media, n > 1 n is a unitless ratio

Some Indices of Refraction

Example 1 What is the apparent speed of light (589 nm) in diamond? Given: n d = 2.42; c = 3.00 x 10 8 m/s Find: v

Example

More About Refraction Wavefront incident on surface Bend as it crosses boundary due to speed change Wave at B reaches D in t traveling at speed v i Transmitted portion of wavefront traveling at speed v i reaches E in same time If n t > n i, v t < v i and AE < BD the wavefront bends Incident medium Transmitting medium nini ntnt B D A E ii tt vitvit vttvtt

Snells Law of Refraction BD = v i t and AE = v t t Multiply both sides by c; n i =c/ v i and n t =c/ v t Which is the Law of Refraction, or Snell's Law The incident, reflected, and refracted rays all lie in the plane-of-incidence Both angles are measured with respect to the normal n i sin i = n r sin r

Snells Law of Refraction If n i > n r the light ray will bend AWAY from the normal If n i < n r the light ray will bend TOWARD the normal

Refraction Details, 1 Light may refract into a material where its speed is lower The angle of refraction is less than the angle of incidence The ray bends toward the normal

Refraction Details, 2 Light may refract into a material where its speed is higher The angle of refraction is greater than the angle of incidence The ray bends away from the normal

Example 2 Suppose that the beam of light in the figure travels in air before it is incident on the glass plate (n g = 1.5) at 60 o. (a) At what angle will it be transmitted into the block? nana ngng nana 1 2 i1 i2 t1 t2 (b) Show that the beam emerges from the far side parallel to the original incoming light Given: air-glass interface (n g = 1.5) and i = 60 o Find: t1 and t2

Example 2 Solution: (a) apply Snells law at the first interface: n a sin i1 = n g sin r1 To two figures

Example 2 (b) apply Snells law at the second interface: n g sin i2 = n a sin r2 The light emerges parallel to the incident beam

Frequency Between Media As light travels from one medium to another, its frequency does not change Both the wave speed and the wavelength do change The wavefronts do not pile up, nor are created or destroyed at the boundary, so must stay the same

Index of Refraction (Extended) The frequency stays the same as the wave travels from one medium to the other v = The ratio of the indices of refraction of the two media can be expressed as various ratios When we talk about wavelengths and colors, we should refer to vacuum wavelengths

Total Internal Reflection As angle of incidence is larger, transmitted beam bends further away from normal. Transmitted beam grows weaker Reflected beam grows stronger. Total internal reflection can occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction Ray 5 shows internal reflection ntnt nini

Critical Angle A particular angle of incidence will result in an angle of refraction of 90 This angle of incidence is called the critical angle sin C = n t /n i For n i > n t ntnt nini

Critical Angle For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary This ray obeys the Law of Reflection at the boundary Total internal reflection occurs only when light attempts to move from a medium of higher index of refraction to a medium of lower index of refraction

Example 3 Imagine a beam of light traveling in a block of glass for which n g = 1.56. Suppose the light comes to the end of the block and impinges on an air-glass interface. What is the minimum incidence angle that will result in all the light being reflected back into the glass. Given: n i = n g = 1.56 and n t = n a = 1.00 Find: the critical angle

Example 3 Solution: Total internal reflection problem. c = 39.9 o

Fiber Optics An application of internal reflection Plastic or glass rods are used to pipe light from one place to another Applications include medical use of fiber optic cables for diagnosis and correction of medical problems Telecommunications

Dispersion The index of refraction in anything except a vacuum depends on the wavelength of the light This dependence of n on is called dispersion Snells Law indicates that the angle of refraction made when light enters a material depends on the wavelength of the light Since visible light coves a range of wavelengths, each color is refracted slightly differently, and this tends to spread out the with light into a rainbow pattern.

Dispersion For Frequency to be maintained, as velocity decreases SO MUST wavelength Thus, the color we see reflected from a material remains constant because color depends upon the energy emitted, which depends upon the energy of the photon absorbed, which depends upon the photons frequency E = hf

Objects are real Images are not Rays of light travel outward from any point on an object and can be collected together to form the same point on an image

Mirrors Reflect rays from an object Create a Virtual Image one that CANNOT be projected onto a screen or surface Reflected rays must be traces BACKWARDS in order to find the location of the image.

Plane Mirror The rays received by the observer diverge from the image point Image is virtual appears behind the mirror Cannot be projected onto a screen S A P V soso sisi ii rr Image is right-side up As far behind the surface as the object is in front of it A single reflection changes a right-handed system into a left-handed one and vice versa

Example 4 Using the figure to the right, prove that the image in a flat mirror is located the same distance behind the mirror as the object is in front of it; that is, the object distance s o equals the image distance s i Given: s o and s i Find: Show that s o = s i S A P V soso sisi ii rr

Example 4 Solution: Geometrical analysis will be used The exterior angle of the triangle SAP equals i + qr, which equal the sum of the alternate interior angles of that triangle But Therefore, which means that triangle VAS and VAP are congruent by angle-angle-side. So The image of the source is the same perpendicular distance behind the mirror as the object is in front of it.

Example 5 What is the length of the smallest vertical plane mirror in which you can see your entire standing body all at once, and how should iut be positioned? Given: s o = s i Find: The minimum height of the mirror A E L B G H F C H M D I J

Example 5 Solution: A ray from your toe will enter your eye, striking point H such that Triangles CHD and CHB are congruent so A E L B G H F C H M D I J Similarly, if you are to see the top of your head A mirror of length will do the job A mirror with half your height with its upper edge lowered by half the distance between your eye and the top of your head is required.

Thin Lenses A thin lens consists of a piece of glass or plastic, ground so that each of its two refracting surfaces is a segment of either a sphere or a plane Lenses are commonly used to form images by refraction in optical instruments

Thin Lens Terminology Optical Axis an imaginary line running through the center of the lens, usually drawn perpendicular to lens Focal Point the point at which light rays will converge TO or diverge FROM as they pass through a lens Focal Length the distance between the lens and the Focal Point Converging Lens bends rays TOWARDS the Focal Point (CONVEX) Diverging Lens bends rays OUTWARDS FROM the Focal Point (CONCAVE)

Thin Lens Shapes These are examples of converging lenses They have positive focal lengths They are thickest in the middle

More Thin Lens Shapes These are examples of diverging lenses They have negative focal lengths They are thickest at the edges

Images A real image is one in which light actually passes through the image point Real images can be projected on screens A virtual image is one in which the light does not pass through the image point The light appears to diverge from that point Virtual images cannot be projected on screens

Images Images can be Upright or Inverted The image produced, and its position s i, depend upon: The distance/position of the object (s o ) The focal length of the lens (f)

Ray Diagrams A ray diagram can be used to determine the position and size of an image They are graphical constructions which tell the overall nature of the image They can also be used to check the parameters calculated from the lens and magnification equations

Drawing A Ray Diagram First draw a horizontal Optical Axis Draw the lens in as a small vertical line perpendicular to the optical axis Plot the focal point on each side at the appropriate distance from the lens ON the optical axis

Drawing A Ray Diagram To make the ray diagram, you need to know The position of the object The focal length/point Three rays are drawn They all start from the same position on the object The intersection of any two of the rays at a point locates the image The third ray serves as a check of the construction

Ray Diagrams for Thin Lenses Ray diagrams are essential for understanding the overall image formation Three rays are drawn The first ray is drawn through the center of the lens and continues in a straight line The second ray is drawn parallel to the optical axis and then passes through (or appears to come from) one of the focal lengths The third ray is drawn through/from the other focal point and emerges from the lens parallel to the optical axis There are an infinite number of rays, these are convenient

Extended Imagery Find where any two rays from an object point cross, that is the corresponding image point Ray 1 toward center goes straight through Ray 2 enter + lens parallel to center axis, emerges passing through the image focus Ray 3 Pass through the object focus of + lens, emerges parallel to central axis Construction of diagram locate object and image focal points, equal distances (f) on each side of the lens D O E P S2 1 Any point on extended object converges to an image point

Extended Imagery Rays converge toward image image is real Rays diverge image is virtual cannot be project on screen D O E P S2 1 yoyo yiyi

Ray Diagram for Converging Lens, s o > f The image is real The image is inverted

Ray Diagram for Converging Lens, s o < f The image is virtual The image is upright

Ray Diagram for Diverging Lens The image is virtual The image is upright

Image Characteristics Converging Lens If the object is farther from the lens than the focal length (s o > f) the image is REAL If the object is closer to the lens than the focal length (s o < f), -- the image is VIRTUAL Diverging Lens Image is always VIRTUAL Refracted rays must be traced backwards from the lens to find the image

Example 6 A convex lens having a 60cm focal length is placed 100cm from a frog. Where will the image of the frog be located? What type of image will it be? What are YOU lookin at?

Example 8 A cave man figurine is 0.75m in front of a thin positive lens that has a focal length of 0.25m. Draw a ray diagram and describe the image.

Example 7 A light bulb is 20mm in front of a positive lens with a focal length of 70mm. Determine the type and location of the image formed.

Optical Systems Are sets of mirrors, lenses, prisms, etc., that perform useful functions with light rays. We will consider Light-Gathering Power Accuracy of the image resolution Ability to change the image magnification, filtering Ability to record an image The location, type and magnification values for the image

Light-Gathering Power The more light gathered by an optical system, the brighter the image will be. A lens gathers more light than a pin-hole camera A large diameter telescope gathers more light than a small diameter telescope

Resolution Aberrations defects in the image produced Chromatic lenses all tend to deflect/refract red light waves more than violet. This can be corrected with multiple lenses.

Resolution Spherical blurring caused by the shape/curvature of the lens itself Diffraction limitation overcome in larger lenses

Ability to Change Image Image Relay a system of two or more lenses used to magnify an image. The image from the first lens becomes the object for the second lens. Essentially a Telescope! Magnification depends upon the distance between the lenses and their focal lengths

Ability to Record an Image Film records the inverse of the intensity of incoming white light produces a NEGATVE of the image which must be processed. Charged Coupled Devices (CCDs) use tiny sensors to detect/record RBG light and digitize the intensity values of each. Resolution in these images depends upon the density/number of Pixels.

Location, Type and Magnification of an Image

Thin Lens Formula Simplified s i = distance of image from lens s o = distance of object from lens f = focal length of lens

Example 9 We wish to place an object 45 cm in front of a lens and have its image appear on a screen 90 cm behind the lens. What must be the focal length of the appropriate positive lens? Given: s o = 0.45 m and s i = 0.90 m Find: f

Example 9 Solution: Lens problem without radii and index, so the Thin Lens Equation f = +0.30 m

Thin Lens Formula, Extended R 1 is positive (C 1 is to the right) R 2 is negative (C 2 is to the left) All interfaces that bulge to the left have positive radii soso sisi Thin-Lens Equation or Lens Makers Formula

Thin Lens Formula, Extended Let s o ; so 1/ s o 0, and s i f Substitute into Eq. (24.1), the Lensmakers Formula (24.2) Same result for s i (1/ s i 0, and s o f )

Focal Points and Planes s o rays entering lens are parallel Rays leave lens, converge to image point F i, image focal point; f i is the image focal length Object at F o rays emerge from lens in parallel bundle s i F o object focal point; f o object focal length Thin lens, same medium on both sides f i = f o FiFi FoFo fifi fofo

Focal Length of a Converging Lens The parallel rays pass through the lens and converge at the focal point The parallel rays can come from the left or right of the lens R 1 > 0 and R 2 < 0 Converging Lens is a Positive Lens

Focal Points and Planes FiFi FoFo fifi fofo R 1 0, so f is negative Diverging lens negative lens

Focal Length of a Diverging Lens The parallel rays diverge after passing through the diverging lens The focal point is the point where the rays appear to have originated

Example 10 A small point of light lies on the central axis 120 cm to the left of a thin bi-convex lens (one curved outward on both sides) having radii of 60 cm and 30 cm. S o = 120 cm 30 cm 60 cm n = 1.50 Given that the index of refraction of the lens is 1.50, find the location of the resulting image point. As far as the image distance is concerned, does it matter which way the lens is facing, that is, which surface is toward the object? Unless otherwise informed, always assume the surrounding medium is air. Given: s o = 1.20 m, R 1 = 0.60 m, R 2 = -0.30 m, and n l = 1.50. Find: s i

Example 10 Solution: Given the radii and index of a lens along with either the object or image distance, use the Lensmaker's Formula. Because the lens is bi-convex, the second radius encountered by incoming rays from the left is taken as a negative number.

Example 10 s i =0.60 m Image distance is positive image to right of lens on axis Reversing the lens ( R 1 = 0.30 m, R 2 = -0.60 m) does not effect the results It doesnt matter which way a thin lens faces

Example 11 Determine the focal length in air of a thin spherical planar-convex lens having a radius of curvature of 50 mm and an index of 1.50. What, if anything, would happen to the focal length if the lens is placed in a tank of water? Given: s o = 1.20 m, R 1 = , R 2 = -0.050 m, and n l = 1.50. Find: f when n m = 1.00 and 1.33

Example 11 Solution: Given the radii and index of a lens use the Lensmaker's Formula When the lens is surrounded by medium n m replace n l by n l / n m = 1.50/1.33. Reduces lens ability to bring rays into convergence Increases f to +0.39 m f = +0.10m

Conventions for Lenses S o, f o + left of O S i, f i + right of O R + if center point is right of O y o, y i + above optical axis

Signs of Thin Lens Parameters Quantity+- soso Real objectVirtual object sisi Real imageVirtual image fConverging lensDiverging lens yoyo Right-side-up objectInverted Object yiyi Right-side-up imageInverted image MTMT Right-side-up imageInverted image

Magnification Ratio of transverse dimension of image to corresponding dimension of object is transverse magnification or magnification M T D O E P S2 1 yoyo yiyi Alternatively soso sisi Minus sign both object and image distances are positive, but image is inverted magnification is negative

Single Lens An object at infinity forms an image at F i As the source point moves closer to the lens, the image point moves further to the right When the source point is at F o, the rays emerge parallel to the central axis No real images of objects closer in than f As the object gets closer to the lens, the image grows in size and moves away from the lens Closer in than f and the image is virtual can operate as a magnifying glass Concave lenses only form virtual images

Example 12 A 5.00-cm-tall matchstick is standing 10 cm from a thin concave lens whose focal length is -30 cm. Determine the location and size of the image and describe it. Draw an appropriate ray diagram. Given: y o = 0.0500 m, s o = 0.10 m, f = -0.30 m. Find: s i and y i

Example 12 Solution: (a) Given the object distance and focal length, use the Thin Lens Equation Eq. (24.4) s i = -0.075 m or -7.5 cm Image distance is negative to left of lens virtual image

Example 12 (b) +0.75 0.038 m FiFi (c) ray diagram

chapter 17. geometric optics the study of how light behaves as it interacts with different media...

Documents