In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $k$. What will be the intensity at a point where the path difference is $\lambda/4$? ($\lambda$ = wavelength of light)

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} = $ $(\cos 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 60^{\circ} = \frac{1}{2})$

In Young's double-slit experiment,the ratio of intensities of bright and dark fringes is $9$. What is the ratio of the intensities of the sources?

Consider an $YDSE$ that has different slit widths. As a result,the amplitudes of waves from the two slits are $A$ and $2A$,respectively. If $I_0$ is the maximum intensity of the interference pattern,then the intensity of the pattern at a point where the phase difference between the waves is $\phi$ 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: $\cos 180^{\circ} = -1, \cos 30^{\circ} = \frac{\sqrt{3}}{2}$)

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $2\pi$ at point $B$. Find the difference between the resultant intensities at point $B$ and point $A$.