If $\frac{I_1}{I_2} = \frac{16}{1},$ then $\frac{I_{\max}}{I_{\min}} = ?$

If the intensity of each wave in the observed interference pattern in Young's double slit experiment is $I_0$,then for some point $P$ where the phase difference is $\phi$,the resultant intensity $I$ will be:

Consider a two-slit interference arrangement (see figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.

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

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:

In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.