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DO NOW:

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**Do Now (2/18/14): a) 0.25 m b) 0.50 m c) 0.75 m d) 1.0 m e) 2.0 m**

Pass in your hw, lab, and Do Now!!! 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 most nearly a) m b) 0.50 m c) 0.75 m d) 1.0 m e) 2.0 m

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Objectives Describe the causes of interference patterns due to diffraction and thin films. Use the diffraction equation. Describe the effects of polarization of light. Determine the wavelength of a light source using a diffraction pattern.

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**Brainstorm: Discuss with your elbow partner (2 min):**

Do you think light is a wave or a particle? Why or why not?

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**Intensity Is the energy it carries per unit of time.**

Proportional to the square of the amplitude of the wave. Color of light is related to it’s λ or f, not intensity. What characteristic of sound is most similar to light intensity?

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**Polarization Polarizing filters**

Only possible if light travels as a wave. I = I0cos2(θ) (Malus’s Law)

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Huygen’s Principle In the 17th Century, Christian Huygens proposed a theory stating: every point on a wave front can be considered as a source of tiny wavelets that spread out in the forward direction at the speed of the wave.

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Huygen’s Principle Every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the waves.

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Huygens’ Principle Other scientists immediately understood that Huygens’ principle predicted that all waves should then spread into the shadow behind an obstacle. We call this bending of waves diffraction. If light is a wave, it should undergo the process of diffraction, and should also be able to undergo the process of interference.

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**Light: Particle or Wave?**

A light source illuminating a single slit

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**Light: Particle or Wave?**

Particle theory prediction for the pattern produced by two slits, side by side:

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**Light: Particle or Wave?**

Wave theory prediction for the pattern produced by two slits, side by side: Note: to easily observe light-wave interference, light from the two sources would have to be coherent and monochromatic.

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**Young’s Double Slit Experiment**

Young saw an interference pattern. To explain this alternating pattern of bright and dark fringes, he understood that at any point on the viewing screen (other than at the center), the two rays of light would have to travel different distances to arrive at the screen. x d ϴ L Path Difference

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**Young’s Double Slit Experiment**

If this path length difference is an integer multiple of the light’s wavelength, a bright fringe is seen. A dark fringe is seen anytime the path length difference is ½ of the λ. d ϴ dsinϴ

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Fringes Diffraction causes fringes:

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Example Problem 24-1 520nm light falls on a pair of narrow slits separated by 0.2mm. How far apart are the fringes near the center of the pattern on a screen 2.5m away? x = 0.064m or 6.4mm

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**Small Angle Approximation**

How could we approximate Young’s formula for small angles??

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**Small Angle Approximation**

Note: For small angles (x << L), sinϴ ≈ tanϴ, so you can approximate the distance between the fringes as…

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Diffraction Gratings A special device called a diffraction grating consists of many equally-spaced parallel lines scratched into a glass plate. The spaces between the scratches act as a source of light, and interference is observed. The interference equation predicts the location of maxima in the interference pattern. The maxima are much thinner & more defined than the pattern created by a double slit. The only difference about these problems (compared to old diffraction problems) will be that you won’t be told the value of d. Instead, you’ll be told the number of lines per distance, which is really 1/d.

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Review:

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Huygens’ principle Wave theory of light – every point on a wave front can be considered as a source of tiny wavelets that spread out in the forward direction at the speed of the wave itself. predicts waves bending around openings

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**Diffraction Young’s Double slit experiment – wave nature of light**

dsinθ = m - constructive interference xm= (m L)/d

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**Plane sound waves of wavelength 0**

Plane sound waves of wavelength 0.12 m are incident on two narrow slits in a box with nonreflecting walls, as shown above. At a distance of 5.0 m from the center of the slits, a first order maximum occurs at point P, which is 3.0 m from the central maximum. The distance between the slits is most nearly a) m b) 0.09 m c) 0.16 m d) 0.20 m e) 0.24 m

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**Single Slit Diffraction**

It turns out that an interference pattern is still observed even when there’s only one point as a source of light. According to Huygens’ principle, each portion of the slit acts as a source of waves. Therefore, light from one portion of the slit can interfere with light from a different portion. Numerical Approximation

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**Single Slit Diffraction**

The only thing that changes about problems with single-slit diffraction is that the interference equation now predicts the location of minima (where the diffraction pattern has minimum intensity), for integral m-values.

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**Different Diffraction Patterns**

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**Diffraction Patterns from Edges**

Diffraction patterns can arise anytime light bends when passing around edges of an obstacle. Shadows of objects therefore contain diffraction patterns, with a bright spot at their centers. You should also be aware that light wave diffraction isn’t observed as much in the macroscopic (big) world because diffraction effects are more pronounced when the size of the opening through which the wave passes is close to the size of the wavelength of the wave.

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Do Now (2/19/14): If the distance between two slits is mm and the distance to a screen is 2.5 m, find the spacing between the first and second order bright fringes for light of wavelength 600 nm. What color is this light?

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**Single Slit Diffraction**

Diffraction of light by a slit of narrow width a

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Example: Light of wavelength 580 nm is incident on a slit of width 0.3 mm. The observing screen is placed 2 m from the slit. Find the positions of the first dark fringes and the width of the central bright fringe.

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Brewster’s Law:

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Polarizing Angle The angle of incidence that satisfies Brewster’s Law.

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**Thin-Film Interference**

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**Thin Film Interference**

When light encounters the boundary between two substances with different indices of refraction , some of the light will be reflected and some will be transmitted. Interference happens. Constructive interference happens when the 2nd wave exits the whole mess in phase with the 1st wave. Depth of film = 1/2λ Air (n=1.00) Oil (n=1.28) Water (n=1.33)

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**Thin Film Interference**

Destructive interference happens when the 2nd wave exits the whole mess out of phase with the 1st wave. Depth of film = 1/4λ Air (n=1.00) Oil (n=1.28) Water (n=1.33)

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**Thin-Film Interference**

When light is reflected upon trying to enter a substance with a higher n-value, it is also shifted by 180°, which is equal to a path difference of 1/2λ. Air (n=1.00) Glass (n=1.56) Glass (n=1.56) Air (n=1.00)

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**Thin-Film Interference**

Constructive interference will happen when the second wave undergoes a total phase change of λ. But since during reflection it undergoes a phase change of 1/2λ when it reflects from the top soap layer, it only needs to undergo a phase change of 1/2λ as it travels the thickness of the film (twice). Air (n=1.00) Soap (n=1.28) Air (n=1.00)

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**Thin Film Interference**

Constructive: Destructive:

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**Thin Film Interference**

The wavelength of light, λn, in a medium with index of refraction n is: Where λ is the wavelength of light in free space

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Example: Calculate the minimum thickness of a soap-bubble film (n=1.33) that will result in constructive interference in the reflected light if the film is illuminated by light with a wavelength in free space of 602 nm. 113 nm

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AP Practice! Try to finish the first two problems before the end of today.

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**Lab: How wide is a human hair?**

A human hair can act just like a double slit. The light going around both edges will bend/diffract and create an interference pattern. This lab will include a writeup

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**Electromagnetic Waves and Optics**

AP PHYSICS UNIT 11 GIANCOLI CH

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**Electromagnetic Waves**

We already know that a changing B-field or flux will produce an electric field (i.e. causes the movement of charges or current) Conversely, James Maxwell came up with the idea that a changing electric field can produce a magnetic field.

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**Electromagnetic Waves**

Accelerating charge gives rise to E&M waves that can even travel through a vacuum. The oscillating electric and magnetic fields are perpendicular to one another. E&M waves move through a vacuum at c = 3.00x108 m/s.

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**Electromagnetic Waves**

The wave-speed equation still applies:

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Speed of Light (c) Ole Roemer first determined that the speed of light was finite. He found that the period of Io, one of Jupiter’s moons, varied slightly depending on the relative motion of Earth and Jupiter. If the Earth was moving away from Jupiter during Io’s orbit the light would have to travel a longer distance, increasing I0’s apparent orbital period.

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**Electromagnetic Spectrum**

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**Inverse Square Relationship**

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**Read sections 22.1 – 22.2 (no math in either), 22.3-22.4 and 22.7**

Ch.22 Homework Read sections 22.1 – 22.2 (no math in either), and 22.7 Questions: 1, 3, 5, 10 & 13 Problems: 5-6, 9 & 16 Due: Tomorrow

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**Ch.22 Homework Answers 5. 1.88E10 Hz 6. 1.008E-10 m 9. 8.33min**

16. radio hears 0.14s soon

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**Physics of Sight We see an object in one of two ways:**

The object is a source of light (sun, fire, light bulb filament) The object reflects light Reflected Light rays scatter from each point on an object. Our brains construct the image of an object assuming that the light entering our eyes travels in straight lines.

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Law of Reflection The angle of reflection is equal to the angle of incidence. ϴ1 ϴ2 ϴ1 ϴ2

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Plane Mirrors The image formed by a plane mirror is a virtual image (cannot be projected onto a screen)

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Spherical Mirrors Concave mirrors reflect incoming parallel light rays so that they pass through a common focal point Convex mirrors reflect incoming parallel light rays so that they appear like they are coming from a focal point behind the mirror.

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**Spherical Mirrors C: Center of curvature F: focal point**

r: radius of curvature f: focal length C F F C f

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Spherical Aberration Technically speaking spherical mirrors do not focus the rays perfectly. And the more spherical a mirror is, the more the image will appear blurred. This defect is called spherical aberration. For very sensitive applications parabolic mirrors are used.

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**Concentrating Solar Power (CSP) Plants**

The sun’s rays are focused on pipe filled a fluid to collect the energy in order to generate electricity.

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**Drawing Ray Diagram Rules**

Rule #1: Draw a ray going out from the object parallel to the principal axis that reflects back through the focal point.

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**Drawing Ray Diagram Rules**

Rule #2: Draw a ray that goes through the focal point (or in a direction like it came from the focal point) then reflects back parallel to the axis.

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**Drawing Ray Diagram Rules**

Rule #3: Draw a ray from the object through the center of curvature. This ray will strike the mirror at a right angle and will reflect back on itself.

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**r >object distance > f**

focal point is + Image is… Real, larger & inverted si (+) so (+)

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**object distance < f so (+) si (-) Image is…**

Virtual, larger & upright

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**Ray Diagram for Convex Mirror**

Image is always… Virtual, smaller & upright so (+) si (-) focal point is (-) Uses: rear view mirror, convenience store mirror

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**The Mirror Equation h0 is always (+)**

hi is (+) if upright, (-) if inverted with respect to the object (+)si image in front of mirror (-) si image behind mirror f (+) for concave mirrors, (-) for convex mirrors

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Example Problem 23-4 A 10cm-tall object is placed 12cm in front of a convex mirror that has a radius of curvature of 35cm. Completely describe the reflected image. (What is its location? Its height? Is it real or virtual? Upright or inverted?) si = -7.1cm, hi = 5.92cm, virtual and upright

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**Example Problems 23-4 and 23-5**

A concave mirror with a radius of curvature of 14cm is used to focus the Sun’s rays. Where are the rays focused, relative to the surface of the mirror? You are standing 3.0m from a convex security mirror in a store. You estimate the height of your image to be half of your actual height. Estimate the focal point of the mirror. 7cm from the surface of the mirror at the focal point f = -3m

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Refraction All lenses redirect light rays by the process of refraction: n1 n2 ϴ2 ϴ1

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Lenses Lenses can be grouped into two main categories: converging and diverging lenses.

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**Ray Diagrams for Lenses…**

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Example Problem Use a ray diagram to show the image height and position for the given object. F F

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**Thin Lens Calculations**

The mirror equation still applies, but now it’s called the thin lens equation. (The magnification equation is still the same, too.) Sign conventions for using the equations are somewhat different now… s0 is (+) if it is on the same side of the lens as the incoming light… which is most of the time. If si is (+) then the image is on the opposite side of the incoming light, (-) if on the same side. f is (+) for converging lenses, (-) for diverging lenses.

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Multiple Lenses If two thin lenses in a row are used to form an image, first find the image of the first lens alone. Then the light approaches a 2nd lens as if it had come from the image. This means that the image formed by lens #1 becomes the object for lens #2. F F F F

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Multiple Lenses The magnification for multiple lenses is just the product of the individual magnifications of lens #1 & #2. (Example) Two converging lenses, each of focal length 20cm, are place 50cm apart from one another, and an object is place 10cm to the left of the first lens. Where is the final image formed, and what is the magnification of the entire system?

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