Chapter 7: Geometrical Optics

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1 Chapter 7: Geometrical Optics 7. Reflection at a Spherical Surface L.O 7.. State laws of reflection Laws of reflection state: L.O The incident ray, the reflected ray and the normal all lie in the same plane. The angle of incidence, i equal the angle of reflection, r as shown in figure below Sketch and use ray diagrams to determine the characteristics of image formed by spherical mirrors L.O 7..3 Use for real object only A spherical mirror is a reflecting surface with spherical geometry. Two types: i. Convex, if the reflection takes place on the outer surface of the spherical shape. ii. Concave, if the reflecting surface is on the inner surface of the sphere. C : centre of curvature of the surface mirror. P : centre of the surface mirror (vertex or pole). Line CP : principal or optical axis. 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. BP3 FYSL

2 Ray diagrams for spherical mirrors: Ray diagram is defined as the simple graphical method to indicate the positions of the object and image in a system of mirrors or lenses. Table below shows the conditions for ray diagram: Ray Ray 2 Ray 3 A ray parallel to the principal axis passes through or diverges from the focal point F after reflection. A ray which passes through or is directed towards the focal point F is reflected as a ray parallel to the principal axis. A ray which passes through or is directed towards the centre of curvature C will be reflected back along the same path. At least any two rays for drawing the ray diagram. BP3 2 FYSL

3 Concave Mirror Object Distance u Ray Diagram Image Property u > r Real Inverted Diminished Between point C and F u = r Real Inverted Same size Formed at point C (Application: Overhead projector) f < u < r Real Inverted Magnified Formed at a distance greater than C u = f Real or virtual Formed at infinity (Application: Torchlight, car spotlight) u < f Virtual Upright Magnified Formed at the back of the mirror (Application: Make-up mirror, shaving mirror, mirror used by dentist) BP3 3 FYSL

4 u at infinity Real Inverted Diminished Formed at point F (Application: Reflection telescope) Convex Mirror Object Distance u Any position in front of the convex mirror Ray Diagram Image Property Virtual Upright Diminished Formed at the back of mirror (Application: rear-mirror and side-mirrors of cars, at sharp corners of a road, and the corners in a supermarket because it has a wide field of view and providing an upright image.) Equation (formula) of spherical mirror: f u v Relationship between focal length, f and radius of curvature, r: ΔFCM is isosceles (FC = FM). Consider ray AM is paraxial (parallel and very close to the principal axis). FM FP FP CP 2 r f 2 For spherical mirror only BP3 4 FYSL

5 Linear magnification of the spherical mirror, m is defined as the ratio between image height, h i and object height, h o. m h h i 0 v u m is a positive value if the image formed is upright and it is negative if the image formed is inverted. Table below shows the sign convention for spherical mirror s equation: Physical Quantity Positive sign (+) Negative sign (-) Object Distance, u Image Distance, v Real object (in front of the mirror) Real image (same side of the object) Virtual object (at the back of the mirror) Virtual image (Opposite side of the object) Focal length, f Concave mirror Convex mirror Linear magnification, m Upright image Inverted image Object/ Image Height, h Upright image Inverted image Note: Real image is formed by the actual light rays that pass through the image. Real image can be projected on the screen. Example Question Solution A dentist uses a small mirror attached to a thin rod to examine one of your teeth. When the tooth is.20 cm in front of the mirror, the image it forms is 9.25 cm behind the mirror. Determine a. the focal length of the mirror and state the type of the mirror used b. the magnification of the image BP3 5 FYSL

6 Question An upright image is formed 20.5 cm from the real object by using the spherical mirror. The image s height is one fourth of object s height. a. Where should the mirror be placed relative to the object? b. Calculate the radius of curvature of the mirror and describe the type of mirror required. c. Sketch and label a ray diagram to show the formation of the image Solution A mirror on the passenger side of your car is convex and has a radius of curvature 20.0 cm. Another car is seen in this side mirror and is.0 m in front of the mirror (behind your car). If this car is.5 m tall, calculate the height of the car image. A person of.60 m height stands 0.60 m from a surface of a hanging shiny globe in a garden. a. If the diameter of the globe is 8 cm, where is the image of the b. person relative to the surface of the globe? c. How tall is the person s image? d. State the characteristics of the person s image. BP3 6 FYSL

7 Exercise Question An object is placed 0 cm in front of a concave mirror whose focal length is 5 cm. Determine a. the position of the image. b. the linear magnification and state the properties of the image. Answer: v = 30 cm; m = 3 A concave mirror forms an image on a wall 3.20 m from the mirror of the filament of a headlight lamp. If the height of the filament is 5.0 mm and the height of its image is 35.0 cm, calculate a. the position of the filament from the pole of the mirror. b. the radius of curvature of the mirror. Answer: 4.57 cm; 9.0 cm a. A concave mirror forms an inverted image four times larger than the object. Find the focal length of the mirror, assuming the distance between object and image is m. b. A convex mirror forms a virtual image half the size of the object. Assuming the distance between image and object is 20.0 cm, determine the radius of curvature of the mirror. Answer : 60 mm, -267 mm If a concave mirror has a focal length of 0 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object. Answer: 5 cm, 5.0 cm 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. Answer: 20 cm, 0 cm behind the mirror A.74 m tall shopper in a department store is 5.9 m from a security mirror. The shopper notices that his image in the mirror appears to be only 6.3 cm tall. a. Is the shopper s image upright or inverted? Explain. b. Determine the radius of curvature of the mirror. Answer: u think,.07 m A concave mirror of a focal length 36 cm produces an image whose distance from the mirror is one third of the object distance. Calculate the object and image distances. Asnwer: 44 cm, 48 cm BP3 7 FYSL

8 7.2 Refraction at a Plane and Spherical Surfaces L.O 7.2. State and use the laws of refraction (Snell s Law) for layers of materials with different densities Refraction is defined as the changing of direction of a light ray and its speed of propagation as it passes from one medium into another. Refraction at a plane Laws of refraction state: The incident ray, the refracted ray and the normal all lie in the same plane. For two given media, sin i sin r n 2 constant n or n sin i n2 sin r Snell s Law where n : refractive index of medium (medium containing incident ray) n 2 : refractive index of medium 2 (medium containing refracted ray) n < n 2 (Medium is less dense medium 2) n > n 2 (Medium is denser than medium 2) The light ray is bent toward the normal The light ray is bent away from the normal Special cases: i = critical angle i > critical angle i = 0 (Must be from denser to less dense medium) BP3 8 FYSL

9 Refractive index (index of refraction) sin i Definition is defined as the constant ratio sin r for the two given media. The value of refractive index depends on the type of medium and the colour of the light. It is dimensionless and its value greater than. Consider the light ray travels from medium into medium 2, the refractive index can be denoted by (Medium containing the incident ray) n 2 velocity of velocity of (Medium containing the refracted ray) light in medium v light in medium 2 v If medium is vacuum, then the refractive index is called absolute refractive index, written as: 2 n velocity of velocity of light in vacuum light in medium c v Relationship between refractive index and the wavelength of light: As light travels from one medium to another, its wavelength, changes but its frequency, f remains constant. The wavelength changes because of different material. The frequency remains constant because the number of wave cycles arriving per unit time must equal the number leaving per unit time so that the boundary surface cannot create or destroy waves. By considering a light travels from medium (n ) into medium 2 (n 2 ), the velocity of light in each medium is given by v f and v2 f2 The refractive index can be written as: wavelength of light in vacuum n 0 wavelength of light in medium Other equation for absolute refractive index in term of depth is given by n real depth apparent depth BP3 9 FYSL

10 L.O Use for spherical surface Equation of spherical refracting surface: n n2 n2 n u v r Table below shows the sign convention for spherical refracting surface s equation: Physical Quantity Positive sign (+) Negative sign (-) Object Distance, u Image Distance, v Radius of curvature, r Real object (the object is on the same side of the spherical surface from where the light is coming) Real image (the image is on the opposite side of the spherical surface from where the light is coming) Convex surface (the center of curvature is on the the opposite side of the spherical surface from where the light is coming) Virtual object (the object is on the opposite side of the spherical surface from where the light is coming) Virtual image (the image is on the same side of the spherical surface from where the light is coming) Concave surface (the center of curvature is on the the same side of the spherical surface from where the light is coming) If the refraction surface is flat (plane), r = The equation (formula) of linear magnification for refraction by the spherical surface is given by hi nv m h n u o 2 BP3 0 FYSL

11 Example Question A fifty cent coin is at the bottom of a swimming pool of depth 3.00 m. The refractive index of air and water are.00 and.33 respectively. Determine the apparent depth of the coin. Solution A cylindrical glass rod in air has a refractive index of.52. One end is ground to a hemispherical surface with radius, r = 3.00 cm as shown in figure below. Calculate, a. the position of the image for a small object on the axis of the rod, 0.0 cm to the left of the pole as shown in figure. b. the linear magnification. (Given the refractive index of air, n a =.00) Figure below shows an object O placed at a distance 20.0 cm from the surface P of a glass sphere of radius 5.0 cm and refractive index of.63. Determine a. the position of the image formed by the surface P of the glass sphere, b. the position of the final image formed by the glass sphere. (Given the refractive index of air, n a =.00) BP3 FYSL

12 Exercise Question We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of 4.0 above the horizontal as shown in figure below. Calculate the depth of the pool. (Given n water =.33 and n air =.00) Answer : 5.6 m A light beam travels at.94 x 0 8 m s - in quartz. The wavelength of the light in quartz is 355 nm. a. Find the index of refraction of quartz at this wavelength. b. If this same light travels through air, what is its wavelength there? (Given the speed of light in vacuum, c = 3.00 x 0 8 m s - ) Answer:.55; 550 nm A pond with a total depth (ice + water) of 4.00 m is covered by a transparent layer of ice of thickness 0.32 m. Determine the time required for light to travel vertically from the surface of the ice to the bottom of the pond. The refractive index of ice and water are.3 and.33 respectively. (Given the speed of light in vacuum is m s -.) Answer: s A small strip of paper is pasted on one side of a glass sphere of radius 5 cm. The paper is then view from the opposite surface of the sphere. Find the position of the image. (Given refractive index of glass =.52 and refractive index of air=.00) Answer : cm in front of the concave surface (second refracting surface) A point source of light is placed at a distance of 25.0 cm from the centre of a glass sphere of radius 0 cm. Find the image position of the source. (Given refractive index of glass =.50 and refractive index of air=.00) Answer : 28 cm at the back of the concave surface (second refracting surface) BP3 2 FYSL

13 7.3 Thin lenses L.O 7.3. Sketch and use ray diagrams to determine the characteristic of image formed by concave and convex lenses Thin lens is defined as a transparent material with two spherical refracting surfaces whose thickness is thin compared to the radii of curvature of the two refracting surfaces. Thin lenses are divided into two types: i. Convex (Converging) lens which are thicker at the centre than the edges ii. Concave (Diverging) lens which are thinner at the centre then at the edges Both convex and concave lenses have two refractive surfaces of radii of curvature r and r 2 with their respective centre of curvatures at C and C 2 : Centre of curvature (point C and C 2 ) is defined as the centre of the sphere of which the surface of the lens is a part. Radius of curvature (r and r 2 ) is defined as the radius of the sphere of which the surface of the lens is a part. Principal (Optical) axis is defined as the line joining the two centres of curvature of a lens. Optical centre (point O) is defined as the point at which any rays entering the lens pass without deviation. BP3 3 FYSL

14 Table below shows the conditions for ray diagram: Ray Ray 2 Ray 3 Ray which is parallel to the principal axis will be deflected by the lens towards/ away from the focal point F. Ray passing through the optical centre is un-deflected. Ray which passes through the focal point (convex lens) or appear to converge to the focal point (concave lens) becomes parallel to the principal axis after emerging from lens. At least any two rays for drawing the ray diagram. BP3 4 FYSL

15 Converging Lens Object Distance u Ray Diagram Image Property u > 2f Real Inverted Diminished Between F 2 and 2F 2 at the back of the lens (Application: Camera lens, human eye lens) u = 2f Real Inverted Same size Formed at point 2F 2 at the back of the lens (Application: Photocopier) f < u < 2f Real Inverted Magnified Formed at a distance greater than 2F 2 at the back of the lens (Application: Projector lens, objective lens of microscope) u = f Real or virtual Formed at infinity (Application: Eyepiece of astronomical telescope) u < f Virtual Upright Magnified Formed in front of the lens (Application: Magnifying lens) BP3 5 FYSL

16 u = Real Inverted Diminished Formed at F 2 (Application: Objective lens of a telescope) Diverging (Concave) lens Object Distance u Any position in front of the diverging lens Ray Diagram Image Property Virtual Upright Diminished Formed in front of the lens L.O L.O L.O Use thin lens equation for real object only Use lens maker s equation Use the thin lens formula for a combination of two convex lenses Thin lens equation for real object only: u v f Lens maker s equation: f n n medium r r material 2 where f r r 2 n material n medium : focal length : radius of curvature of first refracting surface : radius of curvature of second refracting surface : refractive index of lens material : refractive index of medium The radius of curvature of flat refracting surface is infinity, r =. BP3 6 FYSL

17 If the medium is air, then the lens maker s equation can be written as f n r r2 Linear magnification of the spherical mirror, m is defined as the ratio between image height, h i and object height, h o. m h h i 0 v u m is a positive value if the image formed is upright and it is negative if the image formed is inverted. Since u v f, the linear magnification equation can be written as u v f v v m f Table below shows the sign convention for both lens maker s equation and thin lens equation: Physical Quantity Positive sign (+) Negative sign (-) Object Distance, u Image Distance, v Radius of curvature, r Real object (the object is on the same side of the spherical surface from where the light is coming) Real image (the image is on the opposite side of the spherical surface from where the light is coming) Convex surface (the center of curvature is on the the opposite side of the spherical surface from where the light is coming) Virtual object (the object is on the opposite side of the spherical surface from where the light is coming) Virtual image (the image is on the same side of the spherical surface from where the light is coming) Concave surface (the center of curvature is on the the same side of the spherical surface from where the light is coming) Focal length, f Converging lenses Diverging lenses Object/ Image Height, h Upright image Inverted image BP3 7 FYSL

18 Combination of Two Convex Lenses Many optical instruments, such as microscopes and telescopes, use two converging lenses together to produce an image. In both instruments, the st lens (closest to the object)is called the objective and the 2 nd lens (closest to the eye) is referred to as the eyepiece or ocular. The image formed by the st lens is treated as the object for the 2 nd lens and the final image is the image formed by the 2 nd lens. The position of the final image in a two lenses system can be determined by applying the thin lens formula to each lens separately. The overall magnification of a two lenses system is the product of the magnifications of the separate lenses. m m m 2 where m : magnification due to the first lens m 2 : magnification due to the second lens BP3 8 FYSL

19 Example Question A biconvex lens is made of glass with refractive index.52 having the radii of curvature of 20 cm respectively. Calculate the focal length of the lens in Solution a. water, b. carbon disulfide. (Given n w =.33 and n c =.63) A person of height.75 m is standing 2.50 m in from of a camera. The camera uses a thin biconvex lens of radii of curvature 7.69 mm. The lens made from the crown glass of refractive index.52. a. Calculate the focal length of the lens. b. Sketch a labeled ray diagram to show the formation of the image. c. Determine the position of the image and its height. d. State the characteristics of the image. An object is placed 90.0 cm from a glass lens (n =.56) with one concave surface of radius 22.0 cm and one convex surface of radius 8.5 cm. Determine a. the image position. b. the linear magnification. BP3 9 FYSL

20 Question An object is 5.0 cm from a convex lens of focal length 0.0 cm. Another convex lens of focal length 7.5 cm is 40.0 cm behind the first. Find the position and magnification of the image formed by Solution a. the first convex lens b. both lenses Exercise Question A converging lens with a focal length of 90.0 cm forms an image of a 3.20 cm tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Find a. the object position from the lens. b. the image position from the lens. Is the image real or virtual? Answer: 54 cm; 27 cm A thin plano-convex lens is made of glass of refractive index.66. When an object is set up 0 cm from the lens, a virtual image ten times its size is formed. Determine a. the focal length of the lens, b. the radius of curvature of the convex surface. Answer:. cm, 7.33 cm The objective and eyepiece of the compound microscope are both converging lenses and have focal lengths of 5.0 mm and 25.5 mm respectively. A distance of 6.0 mm separates the lenses. The microscope is being used to examine a sample placed 24. mm in front of the objective. a. Determine i. the position of the final image, ii. the overall magnification of the microscope. b. State the characteristics of the final image. Answer: -29 mm, -9.9, u think A converging lens with a focal length of 4.0 cm is to the left of a second identical lens. When a feather is placed 2 cm to the left of the first lens, the final image is the same size and orientation as the feather itself. Calculate the separation between the lenses. Answer: 2.0 cm BP3 20 FYSL

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