Slit 1 of a double slit is wider than slit 2, so that the light from slit 1 has an amplitudes three times that of the light from slit 2. Show that equation `I=I_(max) "cos"^2 phi/2` is replaced by the equation,
`I = (I_(max)/4)1+(3cos^2(phi/2))` .

What is the ratio of slit widths when amplitudes of light waves from them have ratio 1 : sqrt(2) ?

What is the ratio of slit widths, when the amplitude of light waves from them have a ratio sqrt2 : 1 ?

Slit 1 of Young's double-slit experiment is wider then slit 2, so that the light from slits are given as A_(2) = A_(3) A_(0) sin (omega + (pi)/(3)) , The resultant amplitude and intensity, at a point where the path difference between them is zero, are A and I, respectively. Then

In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If I_(m) be the maximum intensity the resultant intensity I when they interface at phase difference phi is given by .......

Slit 1 of Young's double-slit experiment is wider then slit 2, so that the light from slits are given as A_(1) = A_(0) sin omega t and A_(2)= 3 A_(0)sin (ω+π / 3), , The resultant amplitude and intensity, at a point where the path difference between them is zero, are A and I, respectively. Then

Two slits in Young's double slit experiment have widths in the ratio 81:1. What is the the ratio of amplitudes of light waves coming from them ?

Two slits in Young's double slit experiment have widths in the ratio 81:1. What is the the ratio of amplitudes of light waves coming from them ?