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Diffraction and Thin Film Interference

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Presentation on theme: "Diffraction and Thin Film Interference"— Presentation transcript:

1 Diffraction and Thin Film Interference
The word “laser” is actually an acronym for what phrase? "Light Amplification by the Stimulated Emission of Radiation."

2 What is meant by “monochromatic?”
Single color or frequency or wavelength What is meant by “coherent?” Waves are in phase

3 Diffraction Review Diffraction: Energy is spread out when wave passes through a narrow aperture or around the edge of a barrier. Proof that light behaves as a wave! The more narrow the gap, the greater the diffraction effect. The wider the gap, the less diffraction. The longer the wavelength, the greater the diffraction effect. The shorter the wavelength, the less diffraction. 4. What type of incident wave & aperture gives the MOST diffraction effect? Long wavelength and narrow gap Note: Diffraction is NOT noticeable unless the wavelength is LONG compared to aperture width.

4 Interference Review What does the Superposition Principle say about energy from 2 waves occupying the same space in the medium? Instead of two waves, you get a third wave whose shape and size depends on the patterns of the original two waves ? ? What are the two basic types of interference? This behavior is proof of the WAVE NATURE of light (particles cannot occupy the same space at the same time, wave energy CAN!

5 6. Where Can Diffraction Occur?
Assuming wavelength is long compared to barrier’s edge Around a penny and other opaque objects At the edge of large opaque objects AND..

6 During Single&Double Slit Diffraction of Light
In A Ripple Tank Water passes through narrow aperture(s) Areas of Destructive Interference are noticeable – these are called “nodal lines” During Single&Double Slit Diffraction of Light When Laser Light Shines Through Narrow Slit(s) Single-Slit The brightest (constructive interference) fringe in the middle is called the “central maxima.” The next brightest on the left and right are the 1st order, then the 2nd order and so on. The dark (destructive interference) areas are called “minima.” Double-Slit For approximations in Fraunhofer & Fresnel diffraction, see:

7 Single Slit Diffraction Light passes through a narrow single slit
General Equation for Diffraction: Where: m is the “order number” m = 0 (central maxima) m = 1,2,3… for successive minima

8 Example – Single Slit 575 nm light passes through a slit of width mm. An observing screen is set up 3.00 m away. Find the position of the first dark fringe. For the 1st dark fringe, m = 1. The spacing is given by: (b) What is the width of the central maxima? For the distance between x1 and x2, just double the answer from part (a) (6.90 mm) = 13.8 mm

9 Double Slit Diffraction
In A Ripple Tank Water passes through 2 apertures Nodal Lines are where the waves meet with _________________ interference and anti-nodal lines are where the waves meet with ________________ interference Destructive Constructive

10 Double Slit Diffraction - Geometry Geometry
When the screen is moved further away, what happens to the distance between POINT O and POINT P? It increases (this is visible on a screen) assume  

11 Explanation of Geometry
When the light waves arrive at the same point far from the slits, they will be out of phase by an amount proportional to the difference in their path lengths, (r2 − r1). We call the phase difference between these arriving waves, delta, δ. For instance, when the path difference is one wavelength, one wave has gone through one complete cycle more than the other, and δ = 2π radians. When the path difference is λ/2, then δ = π radians, and so on. We can express this proportionality in the equation: δ = 2π (r2 − r1)/λ. If the point where the waves meet is far from the slits in comparison to their separation d, the path difference is given: (r2 − r1) = d sinθ

12 Young’s Double Slit Experiment
m - Order number refers to BRIGHT fringe locations, m = 1,2,3,4…. DARK fringe locations m+½

13 Young’s Diffraction Example
Label the order number for the maxima on the fringe pattern In order to double the distance on the screen, x, from the central maxima to the 3rd order maxima, what change can be made in d, slit separation? Here m = 3. We want to double the distance , x. To do this, halve the denominator; d 1 1 Two slits separated by mm produces an interference pattern in which the fifth dark band is located 12.8 cm from the central maxima when the screen is placed a distance of 8.2 meters away. What is the wavelength of light? What is the angle, θ? dsinθ = mλ, so θ = sin-1 [(4.5)(8.7x10-7m) /2.50x10-4 m] ≈ 0.90o Here m = 4.5 and we will solve for λ. λ= xd/mL = (0.128)(2.50x10-4 m)/ (4.5)(8.2). So the wavelength is: 8.7 x 10-7 m.

14 Diffraction Grating - Many Slits
With monochromatic source, only single color in maximum With incandescent or sun source, each order has a spectrum of color No approximations And:

15 Diffraction Gratings & Emission Spectrum
Substances can be heated to radiate light with identifiable “colors.” By dividing the colors into the visible spectrum with a diffraction grating, the substance can be identified.

16 Diffraction Grating therefore: d = meters # lines
# lines/cm = 1/distance between slits in centimeters therefore: d = meters # lines Spectral Lines Light Source Monochromatic Green Light “White” Light

17 Diffraction Grating Example
Laser light is passed through a diffraction grating with 7000 lines per centimeter. Light is projected onto a screen far away. An observer by the diffraction grating observes the first order maximum 25° away from the central maximum. What is the wavelength of the laser? There are 1/7000 centimeter per line; thus, the distance between lines is 1.4 × 10–4 cm, or 1.4 × 10–6 m. θ is 25° for the first-order maximum, where m = 1. Plugging into dsinθ=mλ, you get a wavelength of just about 6 × 10–7 m, also known as 600 nm. b. If the first order maximum is 40 cm away from the central maximum on the screen, how far away is the screen from the diffraction grating? This is a geometry problem. tan 25° = (40 cm)/L so L = 86 cm. c. How far, measured along the screen, from the central maximum will the 2nd order maximum be? d sin θ = mλ; solve for θ using m = 2. So θ= 2(6.0×10–7 m)/(1.4×10–6 m) = 59o Using Geometry: tan 59° = x/(0.86 m), so x = 143 cm

18 Comparing Diffraction Patterns
Widest Central maxima? Single Slit Most regular pattern? Diffraction Grating Most Intensity Drop Off? Single Slit Best used for identifying substances? Diffraction Grating (look at emission spectrum)

19 Thin Film Interference
Recall Light will shift phase when reflecting off a more dense medium

20 Equations on Thin Films
If n1 > n2, then no phase change. If n1 < n2, then a phase shift of 180 If one surface causes phase shift , then: destructive interference and constructive interference If both surfaces cause phase shift, then switch!

21 Thin Film Interference Newton’s Rings
Only Reflected Ray 2 has a phase shift For Constructive Interference: For Destructive Interference:

22 Thin Film Interference - Double Glass Plates
Only Reflected Ray 2 has a phase shift So constructive, use m+1/2 and destructive, use m Is the fringe adjacent to “t” bright or dark? This is the “mth” fringe Use 2t = (m+ ½)λ to solve for m If it is dark, then m will be a whole integer If it is bring, then m will be a half integer

23 Thin Film Interference – Gas or Oil
Gasoline (n=1.4) floating on water (n=1.33) so both surfaces cause phase shift For Constructive Interference: For Destructive Interference: air gas water Thin film coatings are used to make anti-reflective surfaces for glass. The minimum thickness is “t” t = λ / 4n Where n=index of refraction

24 Thin Film Interference Soap Bubbles & 2 Glass Plates
Both Rays have a phase shift For Constructive Interference: For Destructive Interference: Light partially reflects off 1st surface and then again off the 2nd surface


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