Download presentation

Presentation is loading. Please wait.

Published bySophie Barefield Modified over 4 years ago

1
Light and Optics Chapters 22 and 23 By Aaron Jones

2
Introduction and Reflection Light is recognized as having the properties of a wave and a particle Energy of a light particle is E=hf h=6.63x10 -34 Js (Plank’s Constant) Ray approximation is used to represent beams of light Reflection: when light encounters a a boundary leading into a second medium part of the incident light ray is bent back into the first medium Two types of reflection: specular reflection (reflection of light from a smooth surface into one direction), diffuse reflection (reflection from a rough surface when light is bent in multiple directions The angle of incidence equals the angle of reflection: Ө1’= Ө1

3
Refraction Refraction: When part of a light ray travels from one transparent medium to another transparent medium, and is bent when entering the second medium The angle of refraction determined by Snell’s Law: n 1 sinӨ 1 =n 2 sinӨ 2 Path of light ray through a refracting surface is reversible Index of Refraction: n=c/v λ n = o /n Wavelength in medium. Frequency of wave does NOT change! Dispersion: Light of different wavelengths is bent at slightly different angles when incident on a refracting material (example: prisms, rainbows)

4
More about refraction Total internal reflection: can occur when n 2 <n 1. If angle of incidence exceeds a critical angle, no light is transmitted across the interface. All light gets reflected back into the original medium. Critical angle: The angle of incidence that makes the refracted light ray move parallel to the boundary of the two mediums (Ө c ) sinӨ c =n 2 /n 1 n 1 >n 2

5
Mirrors/Lenses Di= image distance do= object distance f=focal length (all distances measured from the lens) Lens-mirror formula: 1/f = 1/di + 1/do Real image: when the light intersects at an image point, di is positive Virtual image: When light does not pass through the image point but appears to diverge from that point, di is negative Magnification: M=hi/ho= -di/do –hi=image size, ho=object size –Positive magnification indicates an upright or erect image. (usually virtual) –Negative magnification indicates an upside down or inverted image (usually real)

6
Types of mirrors and lenses Flat Mirror: f=infinity, d i = -d o, M=1, image is erect and lifesize Concave Mirrors: reflective surface is “caved in” –Focal length is positive, f=R/2 –Reflected Rays converge –Inverted Real images are created in front of the lens if object distance is greater than the focal length –Erect, Virtual, Magnified images are created when object distance is less than the focal length. Convex Mirrors: Reflective surface is “bowed out” –Focal length and radius of curvature is NEGATIVE, f=R/2 –Reflected rays diverge (spread out) –Only virtual erect images which are reduced in size (M<1) are created. Converging lenses (usually convex surfaces)— –Same as concave mirrors except the light goes through (f is positive) Diverging lenses (usually concave surfaces) – Just like convex mirrors except the light goes through (f is negative)

7
Ray diagrams Geometric way of locating the image created by a mirror or lens. All rays start at the tip of the arrow representing your object Ray#1: Comes in parallel to axis and goes out through a focal point Ray#2: Comes in through a focal point and goes out parallel to optical axis. Ray#3: (Lens) Goes straight through center of lens and doesn’t bend (mirror) Goes through center of curvature, bounces directly back

8
Ray diagram of a concave mirror

9
Lens Aberrations Aberrations: The departure of real (imperfect) images from the ideal image predicted by the theory that all rays reaching the lens or mirror from a point source are focused at single point. Spherical Aberration: –Result from the fact that the focal points of light rays far from the principal axis of a spherical lens or mirror are different from the focal points of light rays passing through the center Chromatic Aberration: –Result from the fact that light rays of different wavelengths focus at different wavelengths focus at different points when refracted by a lens

10
K & C Reflection of light from a smooth surface is called… Specular Reflection

11
K & C The ratio c/v describes what? Index of Refraction (n)

12
K & C Name one of the formulas for Snell’s Law. sinӨ2/sinӨ1=v2/v1=constant Or n1sinӨ1=n2sinӨ2

13
K & C Snell’s Law deals with what the bending of light as it crosses an interface into a new medium. What is this bending called? Refraction

14
K & C What states that all points on a wave front are point sources for the production of spherical secondary waves called wavelets? Huygens’ Principle

15
K & C The critical angle is defined as… The angle of incidence that makes the refracted light ray move parallel to the boundary of the two mediums OR The angle beyond which an incident ray will experience total internal reflection

16
K & C A ______ image is formed when light rays originating from one point on an object reconverge at the image location. This type of image is generally inverted. It can be projected onto a screen. real

17
K & C Magnification is defined as the ratio of… Image height (h’) to object height (h)

18
K & C -The magnification of a lens or mirror is also equal to ______. M = -di/do

19
K & C What formula calculates the focal length of a curved mirror? f=R/2

20
K & C What types of diagrams can be used to determine the image formed by thin lenses? Ray Diagram

21
K & C What is it called when light scatters in all different directions? Diffuse reflection

22
K & C The thin-lens equation is… 1/p + 1/q = 1/f or 1/do + 1/di = 1/f

23
K & C True of False? Magnification for a thin lens is the same formula as Magnification for a mirror. True

24
K & C True or False? The object of the first lens is treated as the object for the second lens in a combination of lenses. False. Image of the first lens is treated as object for the second lens

25
K & C True or False? A chromatic Aberration results from the fact that the focal points of light rays far from the principal axis of a spherical lens or mirror are different from those of rays passing through the center. False. Spherical Aberration

26
K & C Name the three types of mirrors Flat, Concave and Convex

27
K & C The Len’s maker’s equation is… 1/f = (n – 1) (1/R1 – 1/R2)

28
Application A concave mirror has a focal length of 15cm. Find the image distance when the object is located 30cm in front of the mirror. Describe the image that is formed. Solution: 1/do +1/di = 1/f 1/di = 1/f – 1/do di = (1/f – 1/do) -1 di = (1/15cm – 1/30cm) -1 di = 30cm Because the image distance is positive it is REAL, M=-di/do=-1, so we can also describe it as life size and inverted.

29
Application Light (589nm in vacuum) passes through a diamond with an index of refraction of 2.419. What is the speed of light in a diamond? What is the FREQUENCY of the light inside the diamond? What is the wavelength? Solution: n = c/v v = c/n v = (3.00x10^8m/s)/(2.419) v = 1.240x10^8m/s

30
Application If the wavelength of light in a vacuum is 589nm, and the index of refraction for a diamond is 2.419, then what is the wavelength of light in a diamond? Solution: n = λ0/λn λn = λ0/n λn = (589nm)/(2.419) λn = 243nm

31
Application Find the critical angle of Sodium chloride-Ice boundary when the index of refraction of ice is 1.309, and the index of refraction for sodium chloride is 1.544. What would happen if light traveling in the sodium chloride struck the interface with ice at an angle of 60o from the normal? Solution: sinӨc=n2/n1 = (1.309)/(1.544) = 0.847798 Өc = arcsin(0.847798) = 57.973° All of the light would be reflected back into the NaCl, none would go into the ice

32
Application An object is placed 10cm from a convex mirror with a focal length of 6cm; find the position of the image. Solution: 1/do + 1/di = 1/f 1/do = 1/f – 1/di di = (1/f – 1/do)^-1 di = (1/-6 – 1/10)^-1 di = - 3.75cm Note that since the mirror is convex, the focal length is NEGATIVE. The negative image distance means that the image is a virtual one located inside the mirror. Because |di|<|do| the magnification is less than one.

33
Application A converging lens has an index of refraction of 1.66; the front radius of curvature is 11cm, and the back radius of curvature is -17cm. Find the focal length. Solution: 1/f = (n – 1)(1/R1 – 1/R2) = (1.66 – 1)(1/11 – 1/-17) = 0.09882 f = (0.09882)^-1 f = 10.12cm

34
Application You are 30cm from a concave mirror. The magnification of your face is 1.50 times. What is the focal length of the mirror? Solution: M = -di/do -di = M do di = -45cm 1/f = 1/di + 1/do 1/f = 0.011111 f = 90cm

35
Application What is the magnification of a lens which makes a 3m object appear to be an 8m object? Solution: M = h’/h = 2.67

36
Application Two lenses with focal length x and –x are placed in contact along their optical axis. The combined focal length is? Solution: infinity

37
Application What is the index of refraction of a liquid that changes lights wavelength from 490nm to 429nm? What happens to the frequency? Solution: n = λ0/λn n = 1.142 The frequency of a wave does not change when it enters a new medium.

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google