Two coherent souces of intensity ratio `alpha` intefere. Show that `(I_("max")-I_("mini"))/(I_("max")+I_("mini"))=(2sqrt(alpha))/(1+alpha)`

The interference pattern is obtained with two coherent light sources of intensity ratio n.In the interference pattern, the ratio (I_(max)-I_(min))/(I_(max)+I_(min)) will be

Two slits in Young's experiment have widths in the ratio 1:25 The ratio of intensity at the maxima and minima in the interference pattern (I_(max))/(I_(min)) is .

1f (1)/(sqrt3+sqrt2)=sqrt3+alpha then alpha is

Show that ((sqrt(3))/(2) + (i)/(2))^(5) + ((sqrt(3))/(2) - (i)/(2))^(5) = -sqrt(3)

Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are

Two periodic waves of intensities I_1 and I_2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is :

If alpha, beta and gamma the roots of the equation x^(3) + 3x^(2) - 4x - 2 = 0. then find the values of the following expressions: (i) alpha ^(2) + beta^(2) + gamma^(2) (ii) alpha ^(3) + beta^(3) + gamma^(3) (iii) (1)/(alpha)+(1)/(beta)+(1)/(gamma)