Coherent Sources Wavefront splitting Interferometer Young’s Double Slit

Coherent Sources

Wavefront splitting Interferometer

Young’s Double Slit Experiment

Young’s double slit © SPK

Path difference:

For a bright fringe, For a dark fringe, m: any integer

For two beams of equal irradiance (I0)

Visibility of the fringes (V) Maximum and adjacent minimum of the fringe system

Photograph of real fringe pattern for Young’s double slit

The two waves travel the same distance Therefore, they arrive in phase S S'

The upper wave travels one wavelength farther Therefore, the waves arrive in phase S S'

The upper wave travels one-half of a wavelength farther than the lower wave. This is destructive interference S S'

Young’s Double Slit Experiment provides a method for measuring wavelength of the light This experiment gave the wave model of light a great deal of credibility. Uses for Young’s Double Slit Experiment

Phase Changes Due To Reflection An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling Analogous to a reflected pulse on a string μ1 μ2

Phase shift

Fresnel double mirror P1 P2 © SPK

Problem In a Fresnel mirror the angle between the mirrors a=12’. The distance r= 10 cm and b=130 cm. Find The fringe width on the screen and the number of possible maxima. (b) the shift of the interference pattern on the screen when the slit S is displaced by dl=1 mm along the arc of radius r about the center O. (c) The maximum width of the source slit at which the fringe pattern on the screen can still be observed sufficiently sharp.

Fresnel biprism © SPK

Lloyd’s mirror © SPK

Billet’s split lens © SPK

Wavefront splitting interferometers Young’s double slit Fresnel double mirror Fresnel double prism Lloyd’s mirror

Division of Amplitude

Optical beam splitter

Fringes of equal inclination

nf n1 n2 d

Optical path difference for the first two reflected beams

Condition for maxima Condition for minima

Fringes of equal thickness Constant height contour of a topographial map

Wedge between two plates 1 2 glass glass air D t x Path difference = 2t Phase difference  = 2kt -  (phase change for 2, but not for 1) Maxima 2t = (m + ½) o/n Minima 2t = mo/n Fizeau Fringes

Newton’s Ring Ray 1 undergoes a phase change of 180 on reflection, whereas ray 2 undergoes no phase change R= radius of curvature of lens r=radius of Newton’s ring

Reflected Newton’s Ring

Newton’s Ring

Types of localization of fringes

Interference fringes Real Virtual Localized Non-localized

Localized fringe Observed over particular surface Result of extended source

Non-localized fringe Exists everywhere Result of point/line source

POHL’S INTERFEROMETER Real Non-localized Virtual Localized Refer Hecht for details

Problem The width of a certain spectral line at 500 nm is 2×10-2 nm. Approximately what is the largest path difference for which the interference fringes produces by the light are clearly visible?