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Coherent(laser)light of wavelength 550 x 10 -9 m is incident on two slits 0.008

ID: 1964041 • Letter: C

Question

Coherent(laser)light of wavelength 550 x 10-9m is incident on two slits 0.008 mm apart. These slits become coherent, in-phase sources of cylindrical wave fronts which interfere and produce maxima and minima in intensity,"bright lines", on a screen 70 cm away. Draw and lable figure with symbols L1,L2,R,,d,D and y.

a)Calculate the first three "+y" positions on the screen for intensity maxima relative to the central maximum. Here it is appropriate to use the small angle approximation.

b)At a position y = 3.2 cm on the xcreen, what is the phase difference between the two incident waves?

Explanation / Answer

Given Wavelength of incident light = 550*10-9 m Seperation between two slits d = 0.008 mm                                                = ( 0.008 mm ) ( 0.001 / 1 mm )                                                = 8*10-6 m Distance slits and screen L = 70 cm                                          = ( 70 cm ) ( 0.01 m / 1 cm )                                          = 0.7 m a) Linear positin of bright fringe from central maximum at small angles                  y = L ( m / d ) For m =1                  y = L / d                     = ( 0.7 m ) ( 550*10-9 m ) / ( 8*10-6 m )                     = 0.048 m For m =2                  y = 2L / d                     = 2 ( 0.07 m ) ( 550*10-9 m ) / ( 8*10-6 m )                     = 0.096 m For m =3                 y = 3L / d                     = 3( 0.07 m ) ( 550*10-9 m ) / ( 8*10-6 m )                     = 0.148 m                     = 2 ( 0.07 m ) ( 550*10-9 m ) / ( 8*10-6 m )                     = 0.096 m For m =3                 y = 3L / d                     = 3( 0.07 m ) ( 550*10-9 m ) / ( 8*10-6 m )                     = 0.148 m                     = 3( 0.07 m ) ( 550*10-9 m ) / ( 8*10-6 m )                     = 0.148 m __________________________________________________ b)At a position y = 3.2 cm                           = ( 3.2 cm ) ( 0.01 m / 1 cm )                          =  0.032 m              y = L tan         tan = y / L                 = 0.032 m / 0.7 m              = 2.617 o Path difference           d sin   =  ( 8*10-6 m )  sin 2.617                       = 3.65*10-7 m Phase difference             = ( 2 / )  d sin                  = 2 (  3.65*10-7 m ) /  ( 550*10-9 m )                = 1.32                = 4.169 o