In Young's double-slit experiment,if the ratio of the widths of the two slits is $4:9$,then the ratio of the maximum to minimum intensity will be:

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 a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of light used) is $I$. If $I_0$ denotes the maximum intensity,then $\frac{I}{I_0} = $ . . . . . .

In Young's double slit experiment,if the maximum intensity is $I$,then the angular position where the intensity becomes $\frac{I}{4}$ is:

In Young's double slit experiment,one of the slits is wider than the other,so that the amplitude of the light from one slit is double that of the other. If $I_m$ is the maximum intensity,the resultant intensity $I$ when they interfere at a phase difference $\phi$ is given by

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 $\frac{\lambda}{3}$.