Superposition and interference - class-XII

If two points are at the same distance from a source, the light waves arrive at those points with the same path length. Therefore, the path difference is 0, which results in a phase difference of 0.

Points on the same wavefront are in the same phase. Therefore, the phase difference between any two points on the same wavefront is zero, regardless of the distance between them.

The distance x of the nth dark fringe in a wedge-shaped film is given by x = n * lambda / (2 * mu * theta). Substituting the given values for wavelength, refractive index, and wedge angle gives 0.2 mm.

Resultant intensity I_res = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). Given I1 = I, I2 = 9I, phi = pi. cos(pi) = -1. I_res = I + 9I + 2 * sqrt(9 * I^2) * (-1) = 10I - 6I = 4I.

I_res = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). I1 = I, I2 = 4I. For phi = pi/2, cos(pi/2) = 0, so I_res = I + 4I = 5I. For phi = pi, cos(pi) = -1, so I_res = I + 4I + 2 * sqrt(4 * I^2) * (-1) = 5I - 4I = I.

Given that,

Path difference $x=\dfrac{\lambda }{6}$

We know that,

  $ I={{I} _{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) $

 $ \dfrac{I}{{{I} _{0}}}={{\cos }^{2}}\left( \dfrac{\phi }{2} \right)....(I) $

Now, the phase difference is

  $ \phi =\dfrac{2\pi }{\lambda }\times x $

 $ \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6} $

 $ \phi =\dfrac{\pi }{3} $

Now, put the value of $\phi $ in equation (I)

  $ \dfrac{I}{{{I} _{0}}}={{\cos }^{2}}{{30}^{0}} $

 $ \dfrac{I}{{{I} _{0}}}=\dfrac{3}{4} $

 $ \dfrac{I}{{{I} _{0}}}=0.75 $

Hence,
the value of ratio of the intensity at the point is $0.75$ 

The path difference at a point directly in front of one slit is the difference in distances from the two slits. Using the Pythagorean theorem, the distance from the far slit is sqrt(b^2 + d^2) and from the near slit is b. Path difference = sqrt(b^2 + d^2) - b. Using binomial expansion for d << b, sqrt(b^2 + d^2) - b = b(1 + d^2/b^2)^(1/2) - b approx b(1 + d^2/2b^2) - b = d^2 / 2b.

y1 = 10^-6 sin(100t + x/50 + 0.5). y2 = 10^-6 cos(100t + x/50) = 10^-6 sin(100t + x/50 + pi/2). The phase of y1 is phi1 = 100t + x/50 + 0.5. The phase of y2 is phi2 = 100t + x/50 + pi/2. Phase difference = |phi2 - phi1| = |pi/2 - 0.5| = |1.57 - 0.5| = 1.07 radians.

In a YDSE, the central bright fringe occurs where the path difference is zero for all wavelengths, resulting in maximum constructive interference and thus the highest intensity.

When two plane waves make a small angle alpha, the fringe width is given by beta = lambda / alpha. This is derived from the geometry of the interference pattern formed by the two waves.

When two waves y1 = A1 sin(omegat - B1) and y2 = A2 sin(omegat - B2) superimpose, the resultant amplitude A is given by the vector sum of the amplitudes: A = sqrt(A1^2 + A2^2 + 2A1A2*cos(B1 - B2)).

Phase diff. = $\displaystyle\frac{2\pi x}{\lambda}$
Path difference = $\displaystyle\frac{2\pi \times 2.1 \times 10^{-6}}{5460 \times 10^{-10}}$ = 7.692 $\pi$ radian.

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index $\mu $ in time $t$ is given by
$d=vt$            .....(i)