In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda / 6$ is $I'$. If $I_0$ denotes the maximum intensity,then $I'/I_0$ is equal to

In the Young's double slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda / 4$ will be:

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 $\pi / 2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

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:

Two coherent light sources having intensity in the ratio $2x$ produce an interference pattern. Then the value of $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

Waves emitted from two identical sources produce an intensity of $K$ units at a point on the screen. If the path difference between these two waves is $\lambda$,calculate the intensity at a point on the screen where the path difference is $\lambda/4$.