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

The ratio of intensities of two waves is $9 : 1$. They are producing interference. The ratio of maximum and minimum intensities will be

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

Two slits in Young's experiment have widths in the ratio $1 : 25$. The ratio of intensity at the maxima and minima in the interference pattern,$\frac{I_{max}}{I_{min}}$ is

In Young's double-slit experiment using monochromatic light of wavelength $\lambda$,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$?

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?