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CHAPTER 22: Geometrical optics (4 Hours) PowerPoint Presentation
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CHAPTER 22: Geometrical optics (4 Hours)

CHAPTER 22: Geometrical optics (4 Hours)

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CHAPTER 22: Geometrical optics (4 Hours)

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  1. The study of light based on the assumption that light travels in straight lines and is concerned with the laws controlling the reflection and refraction of rays of light. CHAPTER 22: Geometrical optics(4 Hours)

  2. UNIT 22 : GEOMETRICAL OPTICS 22.1 Reflection at a spherical surface 22.2 Refraction at a plane and spherical surfaces 22.3 Thin lenses

  3. At the end of this topic, students should be able to: ( 1 H) • a) State laws of reflection. • b) Sketch and use ray diagrams to determine the characteristics of image formed by spherical mirrors. • c) Use For real object only

  4. The Law of reflection The law of reflection states that • The incident ray, the reflected ray and normal, all lie in the same plane • The angle of incidence i is egual to the angle of reflection r.

  5. 22.1 Reflection at a spherical surface Terms and Definitions • A spherical mirror is a reflecting surface • with spherical geometry. • Two types : • i) convex, if the reflection takes place on • the outer surfaceof the spherical shape. • ii) concave, if the reflecting surface is on • the inner surfaceof the sphere.

  6. 22.1 Reflection at a spherical surface Terms and Definitions Imaginary spherical A A P C P C F F B B f f r r A concave mirror A convex mirror C ~ centre of curvature of the surface mirror. P ~ centre of the surface mirror (vertex or pole). Line CP ~ principal or optical axis.

  7. 22.1 Reflection at a spherical surface Imaginary spherical Terms and Definitions A A P C P C F F B B f f r r AB ~ aperture of the mirror. F ~ focal point of the mirror. f ~ focal length (FP, distance between focal point and the centre of the mirror). r ~ radius of curvature of the mirror.

  8. 22.1 Reflection at a spherical surface Terms and Definitions Focal point, F “point on the principal axis where rays parallel and close to the principal axis pass after reflection”. A concave mirror

  9. 22.1 Reflection at a spherical surface Terms and Definitions F - “ point on the principal axis where rays parallel to the principal axis appear to diverge from after reflection”. A convex mirror Focal point, F

  10. A M P C F f r 22.1 Reflection at a spherical surface Relation between focal length, f and radius of curvature, r is isosceles. (FC=FM) Consider ray AM is paraxial (parallel and very close to the principal axis). FM = FP or FP = 1/2 CP

  11. 22.1 Sketch and Use Ray diagram Images Form by Spherical Mirrors Information about the image in any case can be obtained either by drawing a ray diagram or by calculation using formula. • Ray diagram Ray 1 : A ray parallel to the principal axis is reflected through the focus (focal point).

  12. Images Form by Spherical Mirrors • Ray diagram Ray 2 : A ray passing through the focus is reflected parallel to the principal axis. Ray 3 :A ray passing through the centre of curvature is reflected back through the centre of curvature.

  13. Drawing Compass

  14. Images Form by Spherical Mirrors • Ray diagram

  15. Image formed by concave mirrors • 1) Object beyond C • Between C and F • Real • c) Inverted • d) Smaller than object

  16. Image formed by concave mirrors • 2) Object at C • At C • Real • Inverted • Same size as object

  17. Image formed by concave mirrors 3) Object between C and F a) Beyond C b) Real c) Inverted d) Larger than object (magnified)

  18. Image formed by concave mirrors • 4) Object between F and P • Behind mirror • Virtual • Upright • d) Larger than object • (magnified)

  19. F Notes Image formed by concave mirrors • If the object is at infinity, a real image is • formed at F. Conversely, an object at F • gives a real image at infinity. 5) Object at infinity • At F • Real • Inverted • Smaller than object ii) In all cases, the foot of the object is on the principal axis and its image also lies on this line.

  20. Image formed by a convex mirror Ray 1 :A ray parallel to the axis is reflected as though it came from the focal point. Ray 2 :A ray heading toward the focal point is reflected parallel to the axis. Ray 3 :A ray heading toward the centre of curvature is reflected back on itself.

  21. Image formed by a convex mirror The image always • Virtual • Upright • Smaller than object

  22. b) The mirror equation-calculation using formula A P B M v r u

  23. b) The mirror equation-calculation using formula Object distance = OP = u Image distance = IP = v Radius of curvature = CP = r Object size = OA = h Image size = IB = h’ Focal length = f or

  24. b) The mirror equation-calculation using formula Linear Magnification, m or or

  25. Quantity Positive sign (+) Negative sign (-) Object distance (u) Real object Virtual object Image distance (v ) Real image Virtual image Focal Length (f ) Convex mirror Concave mirror Magnification (m) Inverted image Upright image b) The mirror equation-calculation using formula Sign convention

  26. b) The mirror equation-calculation using formula Example 1.2.1 An object 6 cm high is located 30 cm in front of a convex spherical mirror of radius 40 cm. Determine the position and height of its image. Solution

  27. b) The mirror equation-calculation using formula Example 1.2.2 An object is placed 15 cm from a a) concave mirror b) convex mirror of radius of curvature 20 cm. Calculate the image position and magnification in each case.

  28. b) The mirror equation-calculation using formula Solution 1.2.2 a) Concave mirror , u = +15 cm r = +20 cm f = +10 cm Substituting values and signs in the mirror equation,

  29. b) The mirror equation-calculation using formula Solution 1.2.2 The image is real since v is positive and it is 30 cm in front of the mirror. Magnification, -ve (inverted) The image is twice as high as the object.

  30. b) The mirror equation-calculation using formula Solution 1.2.2 b) Convex mirror, u = +15 cm r = -20 cm f = -10 cm The image is virtual since v is negative and it is 6.0 cm behind the mirror.

  31. b) The mirror equation-calculation using formula Solution 1.2.2 Magnification, The image is two-fifth as high as the object.

  32. b) The mirror equation-calculation using formula Example 1.2.3 What is the focal length of a convex spherical mirror which produces an image one-sixth the size of an object located 12 cm from the mirror ? Solution

  33. b) The mirror equation-calculation using formula Example 1.2.4 When an object is placed 20 cm from a concave mirror, a real image three times is formed. Calculate a) The focal length of the mirror b) Where the object must be placed to give a virtual image three times the height of the object.

  34. b) The mirror equation-calculation using formula Solution 1.2.4 a) u = + 20 cm , m = -3 Using f = +15 cm

  35. b) The mirror equation-calculation using formula Solution 1.2.4 b) Given m = 3 , f = +15 cm Using

  36. b) The mirror equation-calculation using formula Example 1.2.5 An object 2.0 cm high is placed 30 cm from a concave mirror with a radius of curvature of 10 cm. Find the location and its characteristics. Solution Given : h = 2.0 cm, u = +30 cm, r = +10 cm, f = r/2= +5 cm

  37. b) The mirror equation-calculation using formula Solution 1.2.5 Characteristics : 1) smaller 2) in front of the mirror 3) inverted 4) real

  38. b) The mirror equation-calculation using formula Exercise • If a concave mirror has a focal length of 10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.( 15cm, 5.0cm ) • 2) A convex mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image position.( 20cm,10cm behind the mirror )

  39. b) The mirror equation Exercise 3) An object is placed 5.0cm in front of a concave mirror with a 10.0 cm focal length. Find the location of the image and its characteristics.( -10cm, m =-2, virtual, upright and located behind the mirror ) 4) What kind of spherical mirror must be used, and what must be its radius, in order to give an erect image one-fifth as large as an object placed 15 cm in front of it ? (-7.5 cm, convex mirror)

  40. b) The mirror equation Exercise 5) A concave mirror forms an image , on the wall 3.0 m from the mirror, of the filament lamp 10 cm in front of the mirror. a) What is the radius of curvature of the mirror ? b) What is the height of the image if the height of the object is 5 mm ? (19.4 cm, -30)

  41. b) The mirror equation Exercise 6) What are the nature, size, and location of the image formed when a 6 cm tall object is located 15 cm from a spherical concave mirror of focal length 20 cm ? (virtual, upright, -60 cm, + 24 cm) 7) The magnification of a mirror is -0.333. Where is the object located if its image is formed on a card 540 mm from the mirror? What is the focal length ? (1.62 m, +405 mm)

  42. Refraction at aplane and spherical surfaces (1 H) At the end of this chapter, students should be able to: • 22.2.1 State and usethe laws of refraction (Snell’s Law) for layers of materials with different densities. • 22.2.2 Apply for spherical surface.

  43. 22.2 Refraction at a Plane and Spherical Surfaces • Refraction is the change in direction of light • when it passes through matter. 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 enters the second medium. • The ray that enters the second medium is bent • at the boundary and is said to be refracted.

  44. 22.2 Refraction at a Plane and Spherical Surfaces Refraction at a plane surface Normal Reflected ray Incident ray θ1 θ’ θ1 = θ’ Air Glass θ1 > θ2 θ2 transparent medium Refracted ray

  45. 22.2 Refraction at a Plane and Spherical Surfaces Refraction at a plane surface Law of Refraction • The incident ray, refracted ray and the • normal all lie in the same plane. • At the boundary between any two given • materials, the ratio of the sine of the • angle of incidence to the sine of the • angle of refraction is constant for rays • of any particular wavelength (previous • figure). This is known as Snell’s Law.

  46. 22.2 Refraction at a Plane and Spherical Surfaces Refraction at a plane surface Law of Refraction Snell’s Law where v1is the speed of light in medium 1 andv2is the speed of light in medium 2. • This relationship shows that the angle of • refraction Ө2 depends on the speed of • light and on the angle of incidence.

  47. 22.2 Refraction at a Plane and Spherical Surfaces Snell’s Law Refraction at a plane surface nglass > nair Normal Normal v1 > v2 v1 < v2 θ1 v1 v1 θ1 Air Glass Air Glass θ1 > θ2 θ1 < θ2 θ2 v2 v2 θ2 this ray is bent toward normal this ray is bent away from normal

  48. 22.2 Refraction at a Plane and Spherical Surfaces Refraction at a plane surface

  49. CONGRATULATIONS!!! S11T5 S11T5 FIRST RANK (Tutorial Physics UPS2 11/12) S1K1 S1K1 FIRST RANK (Lecture Physics UPS211/12)

  50. 22.2 Refraction at a Plane and Spherical Surfaces Snell’s Law Refraction at a plane surface Index of Refraction, n The speed of light in any material or medium is less than the speed of light in vacuum or air. ..(1.1) • The index of refraction is a dimensionless and never less than 1 (n ≥ 1) because v is usually less than c. • n is equal to unity (n = 1) for vacuum.