Optics Mirrors and Lenses Regular vs Diffuse Reflection

Optics Mirrors and Lenses

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

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 1 )The incident ray, the reflected ray and the normal reflection all lie in the same plane. normal incident ray mirror 2)The incident angel = the reflected angel ˊ reflected ray

24 -1 mirrors

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

example • If the angle of incidence of a ray of light is 42 owhat is each of the following? A-The angle of reflection (42 o) B-The angle the incident ray makes with the mirror (48 o) ident ray and the reflected (90 o)

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. 7

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) 8 What is the distance between the girl and 150 m her image in the mirror? Distance = 150 2 = 300 cm T. Norah Ali Almoneef 150 m

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

24 -2 Thin Lenses

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

Rules For Converging Lenses 1) 2) 3) 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.

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

SS s S S- S-

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

Object Outside 2 F Real; inverted; diminished 1. The image is inverted, i. e. , opposite to the object orientation. 3. The image is diminished in size, i. e. , smaller than the object. 2. The image is real, i. e. , formed by actual light on the opposite side of the lens. Image is located between F and 2 F

Example 3. 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. 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).

Object Between 2 F and F F 2 F 2 F Real; inverted; enlarged F 1. The image is inverted, i. e. , opposite to the object orientation. 3. The image is enlarged in size, i. e. , larger than the object. 2. The image is real; formed by actual light rays on opposite side Image is located beyond 2 F

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.

. 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 ? a b

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.

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

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, . . .

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

Ray Diagrams for Thin Lenses – Diverging For a diverging lens, 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

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

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 R 1 and R 2 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

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 f 1 and f 2 placed next to each other are equivalent to a single lens with a focal length f satisfying

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 For a mirror, parabolic shapes can be used to correct for spherical aberration

Spherical Aberration With SA 37 SA free

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

Multiple lenses can be used to improve aberrations Spherical Aberration Chromatic Aberration

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.

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 1 1 1 + = s’ s f 1 1 1 - 6 = s’ 4 1 P= 0. 04 m f = 4. 0 cm 1 f _ T. Norah Ali Almoneef 1 = s s’ s’ = 12 cm = 25. 0 D M = - 12 / 6 = -2 Negative means real, inverted image 41 1

1 1 1 + = s’ s f 42 T. Norah Ali Almoneef

43 T. Norah Ali Almoneef

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. +20 cm R 1 = 20 cm, R 2 = -40 cm n = 1. 5 f = 80. 0 cm Converging (+) lens.

Example: What must be the radius of the curved surface in a plano-convex lens in order that the focal length be 25 cm? R 1 = , f= 25 cm R 2=? R 1 = 0 f=? R 2 = 0. 5(25 cm) R 2 = 12. 5 cm Convex (+) surface.

Example : What is the magnification of a diverging lens (f = -20 cm) the object is located 35 cm from the center of the lens? First we find q. . . then M F s = +12. 7 cm M = +0. 364

Example 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? 47 Real image, magnification = -1 T. Norah Ali Almoneef

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? 48 T. Norah Ali Almoneef

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. 49 T. Norah Ali Almoneef

24(e). Given a lens with the properties (lengths in cm) R 1 = +30, R 2 = +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 . F 1 50 R 1 R 2 T. Norah Ali Almoneef p . F 2 Image is virtual, upright.

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. T. Norah Ali Magnification = Almoneef +2 51 51

52 T. Norah Ali Almoneef