Consider two coherent monochromatic sources of light of wavelength (lambda) centered at the origin and located...

Consider two coherent monochromatic sources of light of wavelength (lambda) centered at the origin and located at (0, 0,-d) and (0,0,+d). Show that the locus of points, satisfying either the condition for constructive or destructive interference, form a family of hyperboloids of revolution (one for each value of m) with the foci located at the sources. Do this by deriving the equation for the hyperboloid of revolution in Cartesian coordinates (1pt) and locating the focus points (0.5pts). Additionally, what does the surface look like for constructive interference when m=0? Prove your answer (0.5pts). You are not allowed to skips steps by using Wolfram or any other math software (in other words don�t skip important steps).

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Expert Solution

derivation of the equation for the hyperboloid of revolution in cartesian co-ordinates and locating the focus points ::

the surface is the focus of the equation , is given as :

x² / C² + y² / A² - z² / B² = 1 { eq. 1 }

in the three-dimensional cartesian equation. it is quadratic because it is the locus of a second degree equation & each point lies on more than one straight line of the surface.

the particular case of hyperboloid of revolution with C=A for the surface and it is given as :

(x² + y² ) / A² - z² / B² = 1 { eq. 2 }

the surface is also the locus descibed by a hyberbola revolving around its conjugate axis.

the hyperboloid represented by eq. 1 for one sheet and hyperboloid for two sheet can be represented as given ::

x² / a² + y² / b² - z² / c² = - 1 { eq. 3 }

the quantities a, b, c are called the semi-axes of a hyperboloid. Figure (1) shows that hyperboloid of one sheet & another Figure (2) shows for two sheet hyperboloid.

when a = b, the hyperboloid represented by equation are surface of revolution.

for constructive interference,

2d = m

where, m is an integer

when m = 0, then 2d = 0

When two light waves superpose with each other in such away that the crest of one wave coincides the trough of the second wave, then the amplitude of resultant wave becomes zero. Due to destructive interference a dark fringe is obtained on the screen.

genius_generous answered 7 months ago

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