Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

In Young's double slit experiment, the wavelength of light used is $\lambda$. The intensity on the screen at a point for path difference $\lambda/6$ is $X$. The intensity at the point for path difference $\lambda/3$ is (Given: $\cos 180^{\circ} = -1, \cos 30^{\circ} = \frac{\sqrt{3}}{2}$)

In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\frac{I_1 + I_2}{I_0} = $ $(\cos 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 60^{\circ} = \frac{1}{2})$

Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ is

The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the fringe width from the centre is

In a $YDSE$ setup,the intensity due to two coherent beams differs from each other by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima is: