1. Daily Challenge, 1/7 If light is made of waves and experiences interference, how will the constructive and destructive interference appear to us?

2. Light shows Interference Interference is the mutual effect (superposition principle) of two beams of light resulting in a loss of intensity in certain regions (destructive interference) and an increase of intensity of other regions (constructive interference) Double Slit Interference – standing wave pattern similar to water wave tank Interference in thin films – occurs due to reflection off front and back of film being slightly out of phase

3. http://sol.sci.uop.edu/~jfalward/lightinterference/lightinterference.html Young’s Double Slit Experiment Thin Film Interference

4. Diffraction of Light Diffraction – the spreading of light into a region beyond an obstruction (happens on the scale of wavelength) Diffraction Gratings – optical surfaces with many parallel, closely spaced grooves when a beam of monochromatic light is passed through… a principal image, along with less intense 1 st, 2 nd, etc. images, are produced λ = d (sin θ n ) / m λ – wavelength of monochromatic light d – grating constant, in length units m – order number θ n – diffraction angle, measured from principal image

5. Daily Challenge, 1/8 Check out the “rainbow glasses” and “snowflake glasses.” Consider the physics of light discussed yesterday. How do these glasses work?

6. Daily Challenge, 1/9 Minilab:(15 points) Use the diffraction gratings of known spacing to determine the wavelength of a laser. Sketch your experimental setup and show your work. Remember λ = d (sin θ n ) / m λ – wavelength of monochromatic light d – grating constant, in length units m – order number θ n – diffraction angle, measured from principal image

7. Physics Daily Challenge, 1/12 The specifications (accepted values) for groove spacing are 1.6 μm/groove on a CD and 0.7 μm/groove on a DVD. If you use a laser of known wavelength and the diffracted reflection images of the laser beam to calculate a groove spacing of 15,300 Å for a CD, what is the absolute and relative error of your measurement?