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Optics

Optics. Mirrors and Lenses. Regular vs. Diffuse Reflection. Smooth, shiny surfaces have a regula r reflection:. Rough, dull surfaces have a diffuse reflection. Diffuse reflection is when light is scattered in different directions. normal. reflected ray. incident ray. mirror.

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Optics

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  1. Optics Mirrors and Lenses

  2. 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

  3. normal reflected ray incident ray mirror 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.  ˊ 2)The incident angel = the reflected angel

  4. 24-1 mirrors

  5. 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

  6. 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)

  7. 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.

  8. 150 m 150 m 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 Almoneef

  9. 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

  10. 24-2 Thin Lenses

  11. 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

  12. 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.

  13. 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

  14. S- S s S- S S-

  15. 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

  16. F 2F 2F F Object Outside 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

  17. F 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).

  18. F 2F 2F F Object Between 2F and F 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

  19. F 2F 2F 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.

  20. .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

  21. 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.

  22. 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

  23. 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, . . .

  24. 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

  25. 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

  26. 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

  27. Sign Conventions for Thin Lenses

  28. 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

  29. 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

  30. Spherical Aberration With SA SA free

  31. 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

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

  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.

  34. 1 f = 4.0 cm s = 6.0 cm P = = 25.0 D 0.04 m 1 1 1 1 1 1 + = = = 1 1 1 6 s s 4 f f s’ 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’ - s’ = 12 cm s’ M = - 12 / 6 = -2 Negative means real, inverted image T . Norah Ali Almoneef

  35. 1 1 + = 1 s f s’ T . Norah Ali Almoneef

  36. T . Norah Ali Almoneef

  37. -40 cm +20 cm n = 1.5 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. R1 = 20 cm, R2 = -40 cm f = 80.0 cm Converging (+) lens.

  38. R2=? R1= 0 f = ? 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 = , f= 25 cm R2 = 0.5(25 cm) R2=12.5 cm Convex (+) surface.

  39. F 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 s = +12.7 cm M = +0.364

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

  41. 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 Almoneef

  42. 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 Almoneef

  43. . . R1 R2 F1 p F2 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 Image is virtual, upright. T . Norah Ali Almoneef

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