Part V: physical optics Double slit Given: d=.200mm (separation of the slits of

Question

Part V: physical optics

Double slit
Given: d=.200mm (separation of the slits of a double slit)
L=1.00mm (distance between the double slit and a screen onto which the double slit interference pattern is projected.
W=3.00mm (width of the central maximum of the double slit interference pattern)

Part A)
Wavelength of the incident light?
Answer = 9*10^-3m

Answer= 12*10^-7 m

Part B)
Suppose you double the slit separation and keep everything else the same. What is the location of the third constructive interference maximum on neither side of the central maximum?

Explanation / Answer

in    double slit intereference experiment       fringe width       =    mL / d   For cemtral maxima   m    = 1       =   3.0*10^-3 m L = distance between the slit and   the screen    = 1.0*10^- 3 m d   =   slit separation =   0.2*10^-3 m wave length     = d / L   plug all values     we get                 =   0.6*10^-3 m   ----------------------------------------------------------- d ' =    2 (0.2*10^-3 m )   = 0.4*10^-3 m For third maxima   m =3         d'sin   =   3       when    is small   sin   = tan   = y / L    d'   ( y /L )   = 3      here y   = distance from central   to third maxima on either side of the central maxima on the screen and solving for   y    =  3 L   / d'      =   4.5 mm

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