Interference of Light

Which of the following is wrong for interference fringes?

Two coherent sources of intensity ratio $\beta$ interfere. Then the value of $\left(I_{\max }-I_{\min }\right)/\left(I_{\max }+I_{\min }\right)$ is :-

In a Young's double-slit experiment, the intensity of light at each slit is $I_0$. An interference pattern is observed along a direction parallel to the line $S_1{S}_2$ on screen $S$.

The minimum, maximum and the intensity averaged over the entire screen are, respectively,

In Young's double slit experiment, instead of taking slits of equal width, one slit is made twice as wide as the other. Then, in the interference pattern,

Two waves $y_1=A_1\sin \left(\omega t-{\beta }_1\right)$ and $y_2=A_2\sin \left(\omega t-{\beta }_2\right)$ superimpose to form a resultant wave whose amplitude is 

Two light waves are given by, $E_1=2\sin \left(100\pi t-kx+30°\right)$ and $E_2=3\cos \left(200\pi t-kx+60°\right)\text{.}$ The ratio of intensity of first wave to that of second wave is -

There are two coherent point sources of light $S_1$ and $S_2$ in the $x-y$  plane at points $(0,0)$ and $(0,3\lambda )$ respectively. Here, $\lambda$ is wavelength of light. At which one of the following points (given as co-ordinates), the intensity of interference is maximum?

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is $16.$ The intensity of the waves are in the ratio:

In a Young's double slit experiment with light of wavelength $\mathrm{\lambda ,}$ the separation of slits is $d$ and distance of screen is $D$ such that $D\gg d\gg \lambda$ . If the Fringe width is $\beta$ , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

Two coherent point sources $S_1$ and $S_2$ are separated by a small distance $d$ as shown in the figure. The fringes obtained on the screen will be