Optics: Reflection, Refraction Mirrors and Lenses 05/25/2006 Optics: Reflection, Refraction Optics Mirrors and Lenses 24-1 Flat mirrors 24-2 Thin Lenses T.Norah Ali Al-moneef king Saud university 18/05/33 Lecture 16

Regular vs. Diffuse Reflection Smooth, shiny surfaces have a regular reflection: Rough, dull surfaces have a diffuse reflection. Diffuse reflection is when light is scattered in different directions T.Norah Ali Al-moneef king Saud university 18/05/33

Optics: Reflection, Refraction 05/25/2006 Optics: Reflection, Refraction Reflection We describe the path of light as straight-line rays Reflection off a flat surface follows a simple rule: angle in (incidence) equals angle out (reflection) angles measured from surface “normal” (perpendicular) Laws of reflection 1 )The incident ray, the reflected ray and the normal all lie in the same plane. normal incident ray reflected ray mirror  ˊ 2)The incident angel = the reflected angel T.Norah Ali Al-moneef king Saud university 18/05/33 Lecture 16

24-1 Flat mirrors T.Norah Ali Al-moneef king Saud university 18/05/33

An object viewed using a flat mirror appears to be located behind the mirror, because to the observer the diverging rays from the source appear to come from behind the mirror The image distance behind the mirror equals the object distance from the mirror   The image height h’ equals the object height h so that the lateral magnification The image has an apparent left-right reversal The image is virtual, not real! 1- Virtual images - light rays do not meet and the image is always upright or right-side-up“ and also it cannot be projected Image only seems to be there Real images - always upside down and are formed when light rays actually meet T.Norah Ali Al-moneef king Saud university 18/05/33

example If the angle of incidence of a ray of light is 42owhat is each of the following? A-The angle of reflection (42o) B-The angle the incident ray makes with the mirror (48o) C-The angle between the incident ray and the reflected (90o) T.Norah Ali Al-moneef king Saud university 18/05/33

Now you look into a mirror and see the image of yourself. a) In front of the mirror. b) On the surface of the mirror. C)Behind the mirror. T.Norah Ali Al-moneef king Saud university 18/05/33

Example A girl can just see her feet at the bottom edge of the mirror. Her eyes are 10 cm below the top of her head. (a) What is the distance between the girl and her image in the mirror?  Distance = 150  2 = 300 cm T.Norah Ali Al-moneef king Saud university 18/05/33

Signs: Image size and magnification images can be upright (positive image size h’) or inverted (negative image size h’) Define magnification m = h’/h Positive magnification: image orientation unchanged relative to object Negative magnification: image inverted relative to object l m l < 1 if image is smaller than object l m l > 1 if image is bigger than object l m l = 1 if image is same size as object T.Norah Ali Al-moneef king Saud university 18/05/33

24-2 Thin Lenses T.Norah Ali Al-moneef king Saud university 18/05/33

A lens is a transparent material made of glass or plastic that refracts light rays and focuses (or appear to focus) them at a point A converging lens will bend incoming light that is parallel to the principal axis toward the principal axis.Any lens that is thicker at its center than at its edges is a converging lens with positive f. A diverging lens will bend incoming light that is parallel to the principal axis away from the principal axis.Any lens that is thicker at its edges than at its center is a diverging lens with negative f T.Norah Ali Al-moneef king Saud university 18/05/33

Rules For Converging Lenses Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis. An incident ray which passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens. T.Norah Ali Al-moneef king Saud university 18/05/33

T.Norah Ali Al-moneef king Saud university 18/05/33

S- S s S- S S- T.Norah Ali Al-moneef king Saud university 18/05/33

The lens maker’s formula (lens in a medium) ( n lens = index of refraction of the lens material) ( nmedium= index of refraction of the medium) The focal length f for a lens. ( n = index of refraction of the lens material) T.Norah Ali Al-moneef king Saud university 18/05/33

T.Norah Ali Al-moneef king Saud university 18/05/33

Ray Diagram for Converging Lens, S > f The image is real The image is inverted The image is on the back side of the lens 18/05/33 T.Norah Ali Al-moneef king Saud university

T.Norah Ali Al-moneef king Saud university 18/05/33

Ray Diagram for Converging Lens, S < f The image is virtual The image is upright The image is larger than the object The image is on the front side of the lens 18/05/33 T.Norah Ali Al-moneef king Saud university

Object Outside 2F F 2F Real; inverted; diminished 1. The image is inverted, i.e., opposite to the object orientation. 2. The image is real, i.e., formed by actual light on the opposite side of the lens. 3. The image is diminished in size, i.e., smaller than the object. Image is located between F and 2F T.Norah Ali Al-moneef king Saud university 18/05/33

Object at 2F F 2F Real; inverted; same size 1. The image is inverted, i.e., opposite to the object orientation. 2. The image is real, i.e., formed by actual light on the opposite side of lens. 3. The image is the same size as the object. Image is located at 2F on other side T.Norah Ali Al-moneef king Saud university 18/05/33

Object Between 2F and F F 2F Real; inverted; enlarged 1. The image is inverted, i.e., opposite to the object orientation. 2. The image is real; formed by actual light rays on opposite side 3. The image is enlarged in size, i.e., larger than the object. Image is located beyond 2F T.Norah Ali Al-moneef king Saud university 18/05/33

Object at Focal Length F Parallel rays; no image formed When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed. T.Norah Ali Al-moneef king Saud university 18/05/33

Object Inside F Virtual; erect; enlarged 1. The image is erect, i.e., same orientation as the object. 2. The image is virtual, i.e., formed where light does NOT go. 3. The image is enlarged in size, i.e., larger than the object. Image is located on near side of lens T.Norah Ali Al-moneef king Saud university 18/05/33

Example. A magnifying glass consists of a converging lens of focal length 25 cm. A bug is 8 mm long and placed 15 cm from the lens. What are the nature, size, and location of image. F S = 15 cm; f = 25 cm S-= -37.5 cm The fact that S- is negative means that the image is virtual (on same side as object). T.Norah Ali Al-moneef king Saud university 18/05/33

. Example Where must an object be placed to have unit magnification ( M = 1.00) (a) for a converging lens of focal length 12.0 cm ? (b) for a diverging lens of focal length 12.0 cm ? b a T.Norah Ali Al-moneef king Saud university 18/05/33

example A person uses a converging lens that has a focal length of 12.5 cm to inspect a gem. The lens forms a virtual image 30.0 cm away. Determine the magnification. Is the image upright or inverted? solution Since , the image is upright. T.Norah Ali Al-moneef king Saud university 18/05/33

example A ray that starts from the top of an object and runs parallel to the axis of the lens, would then pass through the a)principal focus of the lens b)center of the lens C)secondary focus of the lens T.Norah Ali Al-moneef king Saud university 18/05/33

Example 5: Derive an expression for calculating the magnification of a lens when the object distance and focal length are given. From last equation: = -s M Substituting for q in second equation gives . . . Thus, . . . T.Norah Ali Al-moneef king Saud university 18/05/33

Diverging Thin Lens Incoming parallel rays DIVERGE from a common point FOCAL We still call this the point Same f on both sides of lens Negative focal length Thinner in center T.Norah Ali Al-moneef king Saud university 18/05/33

Ray Diagrams for Thin Lenses – Diverging For a diverging lens, the following three rays are drawn: Ray 1 is drawn parallel to the principal axis and emerges directed away from the focal point on the front side of the lens Ray 2 is drawn through the center of the lens and continues in a straight line Ray 3 is drawn in the direction toward the focal point on the back side of the lens and emerges from the lens parallel to the principal axis T.Norah Ali Al-moneef king Saud university 18/05/33

Ray Diagram for Diverging Lens The image is virtual The image is upright The image is smaller The image is on the front side of the lens 18/05/33 T.Norah Ali Al-moneef king Saud university

Sign Conventions for Thin Lenses Quantity Positive When Negative When Object locatio (s) Object is in front of the lens Object is in back of the lens Image location (sˊ) Image is in back of the lens Image is in front of the lens Image height (h’) Image is upright Image is inverted R1 and R2 Center of curvature is in back of the lens Center of curvature is in front of the lens Focal length (f) Converging lens Diverging lens 18/05/33 T.Norah Ali Al-moneef king Saud university

T.Norah Ali Al-moneef king Saud university 18/05/33

Example : An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram. . . An object placed within the focal point of a converging lens produces a virtual image. Solution: a. The thin lens equation gives = -30 cm and M = 3.0. The image is virtual, enlarged, and upright. b. See the figure. S’ T.Norah Ali Al-moneef king Saud university 18/05/33

Example. Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm from the lens, on the same side as the object? Since the lens is diverging, the focal length is negative. The lens equation gives = 100 cm. S T.Norah Ali Al-moneef king Saud university 18/05/33

T.Norah Ali Al-moneef king Saud university 18/05/33

T.Norah Ali Al-moneef king Saud university 18/05/33

The power of lens The reciprocal of the focal length = the power of lens If the focal length f is measured in meters then ;p measured in diopters if two lenses with focal length f1 and f2 placed next to each other are equivalent to a single lens with a focal length f satisfying T.Norah Ali Al-moneef king Saud university 18/05/33

Spherical Aberration Results from the focal points of light rays far from the principle axis are different from the focal points of rays passing near the axis 18/05/33 T.Norah Ali Al-moneef king Saud university

Spherical Aberration With SA SA free T.Norah Ali Al-moneef king Saud university 18/05/33

Chromatic Aberration Different wavelengths of light refracted by a lens focus at different points Violet rays are refracted more than red rays The focal length for red light is greater than the focal length for violet light Chromatic aberration can be minimized by the use of a combination of converging and diverging lenses 18/05/33 T.Norah Ali Al-moneef king Saud university

Multiple lenses can be used to improve aberrations Spherical Aberration Chromatic Aberration T.Norah Ali Al-moneef king Saud university 18/05/33

Lens Aberrations Chromatic aberration can be improved by combining two or more lenses that tend to cancel each other’s aberrations. This only works perfectly for a single wavelength, however. T.Norah Ali Al-moneef king Saud university 18/05/33

An object is placed 6.0 cm in front of a convex thin lens of focal length 4.0 cm. Where is the image formed and what is its magnification and power? s = 6.0 cm f = 4.0 cm 1 s f = _ 1 s s’ f + = s’ 1 6 4 = - s’ = 12 cm s’ P = 1 0.04 m = 25.0 D M = - 12 / 6 = -2 Negative means real, inverted image T.Norah Ali Al-moneef king Saud university 18/05/33

1 s s’ f + = T.Norah Ali Al-moneef king Saud university 18/05/33

T.Norah Ali Al-moneef king Saud university 18/05/33

Example 1. A glass meniscus lens (n = 1 Example 1. A glass meniscus lens (n = 1.5) has a concave surface of radius –40 cm and a convex surface whose radius is +20 cm. What is the focal length of the lens. -40 cm +20 cm n = 1.5 R1 = 20 cm, R2 = -40 cm f = 80.0 cm Converging (+) lens. T.Norah Ali Al-moneef king Saud university 18/05/33

Example: What must be the radius of the curved surface in a plano-convex lens in order that the focal length be 25 cm? R1= R2=? f = ? R1 = , f= 25 cm R2 = 0.5(25 cm) R2 = 12.5 cm Convex (+) surface. T.Norah Ali Al-moneef king Saud university 18/05/33

Example : What is the magnification of a diverging lens (f = -20 cm) the object is located 35 cm from the center of the lens? F First we find q . . . then M s = +12.7 cm M = +0.364 T.Norah Ali Al-moneef king Saud university 18/05/33

Example Real image, magnification = -1 An object is placed 20 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? Real image, magnification = -1 T.Norah Ali Al-moneef king Saud university 18/05/33

Example An object is placed 8 cm in front of a diverging lens of focal length 4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? T.Norah Ali Al-moneef king Saud university 18/05/33

Example 24(b). Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Image is real, inverted. T.Norah Ali Al-moneef king Saud university 18/05/33

24(e). Given a lens with the properties (lengths in cm) R1 = +30, R2 = +30, s = +10, and n = 1.5, find the following: f, s and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Real side Virtual side R1 . F1 F2 p R2 Image is virtual, upright. T.Norah Ali Al-moneef king Saud university 18/05/33

Example An object is placed 5 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object. Magnification = +2 T.Norah Ali Al-moneef king Saud university 18/05/33 55

T.Norah Ali Al-moneef king Saud university 18/05/33

Summary: Lensmaker’s Equation + - R1 and R2 are interchangeable R1, R2 = Radii n= index of glass f = focal length R1 and R2 are positive for convex outward surface and negative for concave surface. Focal length f is positive for converging and negative for diverging lenses. T.Norah Ali Al-moneef king Saud university 18/05/33

Summary of Math Approach Lens Equation: Magnification: T.Norah Ali Al-moneef king Saud university 18/05/33

Summary of Sign Convention 1. Object p and image q distances are positive for real and images negative for virtual images. 2. Image height y’ and magnification M are positive for erect negative for inverted images 3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors. T.Norah Ali Al-moneef king Saud university 18/05/33

Alternative Solutions It might be useful to solve the lens equation algebraically for each of the parameters: Be careful with substitution of signed numbers!