Chapter 36 - Lenses. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

Transcription

1 Chapter 36 - Lenses A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

2 Objectives: After completing this module, you should be able to: Determine the focal length of converging and diverging lenses. Apply the lensmaker s s equation to find parameters related to lens construction. Use ray-tracing techniques to construct images formed by converging and diverging lenses. ind the location, nature,, and magnification of images formed by converging and diverging lenses.

3 Refraction in Prisms If we apply the laws of refraction to two prisms, the rays bend toward the base, converging light. Two prisms base to base Parallel rays, however, do not converge to a focus, leaving images distorted and unclear.

4 Refraction in Prisms (Cont.) Similarly, inverted prisms cause parallel light rays to bend toward the base (away from the center). Two prisms apex to apex Again there is no clear virtual focus, and once again, images are distorted and unclear.

5 Converging and Diverging Lens If a smooth surface replaces the prisms, a well-defined focus produces clear images. Converging Lens Diverging Lens Real focus Virtual focus Double-convex Double-concave

6 The ocal Length of Lenses Converging Lens ocal length f Diverging Lens f - + f The focal length f is positive for a real focus (converging) and negative for a virtual focus.

7 The Principal ocus Since light can pass through a lens in either direction, there are two focal points for each lens. Left to right The principal focal point is shown here. Yellow is the other one. Now suppose light moves from right to left instead... Right to left

8 Types of Converging Lenses In order for a lens to converge light it must be thicker near the midpoint to allow more bending. Double- convex lens Plano- convex lens Converging meniscus lens

9 Types of Diverging Lenses In order for a lens to diverge light, it must be thinner near the midpoint to allow more bending. Double- concave lens Plano- concave lens Diverging meniscus lens

10 Lensmaker s s Equation The focal length f f for a lens ( n 1) f R R 1 2 The Lensmaker s Equation: Positive (Convex) R 1 R 2 Negative (Concave) R Surfaces of different radius Sign convention

11 Signs for Lensmaker s s Equation R 1 and R 2 are interchangeable R 1 R R 1, R 2 = Radii n= index of glass f = focal length ( n 1) f R R R 11 and R 22 are positive for convex outward surface and negative for concave surface ocal length f f is is positive for converging and negative for diverging lenses.

12 Example 1. A glass meniscus lens (n( 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? R 1 = 20 cm, R 2 = -40 cm ( n 1) f R R cm (1.5 1) f 20 cm ( 40 cm 40 cm +20 cm n = 1.5 f f = 20.0 cm Converging (+) lens.

13 Example 2: 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 =,, R 2 = 25 cm ( n 1) f R (1.5 1) 25 cm R2 R2 0 R 2 =? R 1 = f =? R 2 = 0.5(25 cm) R 22 = 12.5 cm Convex (+) surface.

14 Terms for Image Construction The near focal point is the focus on the same side of the lens as the incident light. The far focal point is the focus on the opposite side to the incident light. Converging Lens ar focus Diverging Lens ar focus Near focus Near focus

15 Image Construction: Ray 1: A ray parallel to to the lens axis passes through the far focus of of a converging lens or or appears to to come from the near focus of of a diverging lens. Converging Lens Ray 1 Diverging Lens Ray 1

16 Image Construction: Ray 2: A ray passing through the near focal point of of a converging lens or or proceeding toward the far focal point of of a diverging lens is is refracted parallel to to the lens axis. Converging Lens Ray 1 Diverging Lens Ray 1 Ray 2 Ray 2

17 Image Construction: Ray 3: A ray passing through the center of of any lens continues in in a straight line. The refraction at at the first surface is is balanced by the refraction at at the second surface. Converging Lens Ray 1 Ray 3 Diverging Lens Ray 1 Ray 2 Ray 2 Ray 3

18 Images Tracing Points Draw an arrow to to represent the location of of an object, then draw any two of of the rays from the tip of of the arrow. The image is is where lines cross. 1. Is the image erect or inverted? 2. Is the image real or virtual? Real images are always on the opposite side of the lens. Virtual images are on the same side. 3. Is it enlarged, diminished, or same size?

19 Object Outside 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 rays in front of mirror. Image is is located between and 2

20 Object at Real; inverted; same size 1. The image is inverted; i.e., opposite to the object orientation. 3. The image is the same size as the object. 2. The image is real; ; i.e., formed by actual light rays in front of the mirror. Image is is located at at 2 on other side

21 Object Between 2 and 2 2 Real; inverted; enlarged 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 the opposite side Image is is located beyond 2

22 Object at ocal Length 2 2 Parallel rays; no image formed When the object is is located at at the focal length, the rays of of light are parallel. The lines never cross, and no image is is formed.

23 Object Inside 2 2 Virtual; erect; enlarged 1. The image is erect; i.e., same orientation as the object. 3. The image is enlarged in size; i.e., larger than the object. 2. The image is virtual; i.e., formed where light does NOT go. Image is is located on near side of of lens

24 Review of Image ormations Parallel Virtual; Real; rays; no erect; inverted; image formed diminished same enlarged size Object Outside 2 Region

25 Diverging Lens Imaging All images formed by diverging lenses are erect, virtual,, and diminished.. Images get larger as object approaches. Diverging Lens Diverging Lens

26 Analytical Approach to Imaging y 2 2 f -y p q Lens Equation: p q f Magnification: y' q M y p

27 Same Sign Convention as or Mirrors 1. Object p and image q distances are positive for real and images negative for virtual images. 2. Image height y and magnification M are positive for erect negative for inverted images p q f y' q M y p 3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors.

28 Working With Reciprocals: The lens equation can easily be solved by using the reciprocal button (1/x( 1/x) ) on most calculators: p q f Possible sequence for finding f on linear calculators: inding f: P 1/x + q 1/x = 1/x Same with reverse notation calculators might be: inding f: P 1/x Enter q 1/x + 1/x

29 Alternative Solutions It might be useful to solve the lens equation algebraically for each of the parameters: p q f p qf q pf q f p f f qp q p Be careful with substitution of of signed numbers!

30 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 the image? p = 15 cm; f = 25 cm p q f q pf p f (15 cm)(25 cm) 15 cm - 25 cm q = cm The fact that q is is negative means that the image is is virtual (on same side as object).

31 Example 3 Cont.) 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 is the size of the image? y y p = 15 cm; q = cm y' q M y p y ' ( 37.5 cm) 8 mm 15 cm Y = +20 mm The fact that y y is is positive means that the image is is erect. It It is is also larger than object.

32 Example 4: What is the magnification of a diverging lens ( (f = -20 cm) ) if the object is located 35 cm from the center of the lens? irst we find q... then M y' q M p q f y p q pf p f (35 cm)(-20 cm) 35 cm - (-20 cm) q = cm M q ( 12.7 cm) M = p 35 cm

33 Example 5: Derive an expression for calculating the magnification of a lens when the object distance and focal length are given p q f q pf p rom last equation: q = -pm f y' q M y p Substituting for q in second equation gives... pf pm p f Thus,... f M p f Use this expression to verify answer in Example 4.

34 Summary A converging lens is is one that refracts and converges parallel light to to a real focus beyond the lens. It It is is thicker near the middle. The principal focus is denoted by the red. A diverging lens is is one that refracts and diverges parallel light which appears to to come from a virtual focus in in front of of the lens.

35 Summary: Lensmaker s s Equation R 1 and R 2 are interchangeable R 1 R R 1, R 2 = Radii n= index of glass f = focal length ( n 1) f R R R 1 and R 22 are positive for convex outward surface and negative for concave surface ocal length f f is is positive for converging and negative for diverging lenses.

36 Summary of Math Approach y 2 2 f -y p q Lens Equation: p q f Magnification: y' q M y p

37 Summary of Sign Convention 1. Object p and image q distances are positive for real and images negative for virtual images. 2. Image height y and magnification M are positive for erect negative for inverted images p q f y' q M y p 3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors.

38 CONCLUSION: Chapter 36 Lenses