#### Transcript d - Madison Public Schools

Concepts and Math! Convex Lenses: Summary Object Location Image Orientation Image Size Image Type Beyond c Inverted Reduced Real At c Inverted Same as object Real Between c and f Inverted Enlarged Real At f No image No image No image Closer than f Upright Enlarged Virtual This is the same as for a concave mirror! This is why convex lenses held close to an object make good magnifying glasses! Concave Lenses: Summary Object Location Image Orientation Image Size Image Type Anywhere Upright Reduced Virtual This is the same as for a convex mirror! Converging lenses and converging mirrors have the same image properties. Diverging lenses and diverging mirrors have the same image properties. Return of the Whiteboard! A large lens is used to focus an image of an object onto a screen. If the left half of the lens is covered with a dark card, which of the following occurs? (A) The left half of the image disappears. (B) The right half of the image disappears. (C) The image becomes blurred. (D) The image becomes dimmer. (E) No image is formed. Although principal rays help guide us to locate the image, we cannot forget the important fact that each point on the object emits rays in all directions. The lens is completely filled with rays from every point of the object! (The image is also formed by infinite rays from the middle of the object, the bottom of the object, etc.) So, if we cover half of the lens... The entire image would still exist! However, less light would be forming the image. Therefore the image would be dimmer. Which three of the glass lenses above, when placed in air, will cause parallel rays of light to converge? (A) I, II, and III (B) I, III, and V (C) l, IV, and V (D) II, III, and IV (E) II, IV, and V I, III, and V are more convex than concave – they will cause light rays to converge. II and IV are more concave than convex – they will cause light rays to diverge. Geometric Optics Equation #1 1 1 1 + = di do f Applies to both mirrors and lenses. di is the distance from the image to the mirror or lens do is the distance from the object to the mirror or lens f is the focal length of the mirror or lens Focal length can be positive or negative! Mirrors and lenses that have a focal point will have a positive focal length. Mirrors and lenses that have an antifocal point will have a negative focal length. Converging lenses and mirrors (concave mirrors and convex lenses) have a positive focal length. Diverging lenses and mirrors (convex mirrors and concave lenses) have a negative focal length. A Point of Possible Confusion • Real images have a positive di • Virtual images have a negative di If you end up with a negative di when you do the calculations, it means that a virtual image is produced! This applies to both mirrors and lenses, and you must be consistent! Whiteboard Problem Solving I A postage stamp is placed 30 centimeters to the left of a converging lens of focal length 60 centimeters. Where is the image of the stamp located? (A) 60 cm to the left of the lens (B) 20 cm to the left of the lens (C) 20 cm to the right of the lens (D) 30 cm to the right of the lens (E) 60 cm to the right of the lens Whiteboard Problem Solving II A concave mirror with a radius of curvature of 1.0 m is used to collect light from a distant star. The distance between the mirror and the image of the star is nearly (A) 0.25 m (B) 0.50 m (C) 0.75 m (D) 1.0 m (E) 2.0 m Solution! A star is so far away that we can comfortably use the approximation do = ∞! This gives 1 1 1 + = di ¥ f Since 1/∞ = 0, this results in di = f Geometric Optics Equation #2 hi di =ho do hi is the height of the image ho is the height of the object A negative image height means that the image is inverted. Memorize both of these equations (don’t forget the negative sign!) Attack of the Whiteboard 15 cm 5 cm f’ f’ 10 cm 10 cm Where will the image be located and what will it look like? do = 15 cm f = -10 cm 1 1 1 + = di 15 -10 di = -6 cm ho = 5 cm hi -6cm =5 cm 15 cm hi = -2 cm f’ (Virtual image 6 cm from lens) (Inverted and reduced image) f’ Deep Whiteboard Thoughts! do ho An object of height h0 is located at a distance do from a plane mirror. 1 1 1 + = Using the mathematical models and di do f hi di =- , ho do determine the location and height of the image formed by the plane mirror! Hint: What is the radius of curvature of a plane mirror? do fplane mirror = ∞ ho 1 1 1 + = = 0! di do ¥ di = -do This means that the mirror will produce a virtual image (negative image distance) that is equidistant from the mirror. hi di =ho do Since di = -do -di / do = 1 Therefore hi / ho = 1 hi = ho The image is the same height as the object!