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 the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to:

In a double slit experiment,the separation between the slits is $d = 0.25 \, cm$ and the distance of the screen $D = 100 \, cm$ from the slits. If the wavelength of light used is $\lambda = 6000 \mathring{A}$ and $I_0$ is the intensity of the central bright fringe,the intensity at a distance $x = 4 \times 10^{-5} \, m$ from the central maximum is

In the Young's double slit experiment,the intensities at two points $P_1$ and $P_2$ on the screen are respectively $I_1$ and $I_2$. If $P_1$ is located at the centre of a bright fringe and $P_2$ is located at a distance equal to a quarter of fringe width from $P_1$,then $\frac{I_1}{I_2}$ is

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

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

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} = $ . . . . . .