In Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ is $I$. If $I_0$ denotes the maximum intensity,find the ratio $\frac{I}{I_0}$.

If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

In a $YDSE$,bichromatic light of wavelengths $\lambda_1 = 400\, nm$ and $\lambda_2 = 560\, nm$ is used. The distance between the slits is $d = 0.1\, mm$ and the distance between the plane of the slits and the screen is $D = 1\, m$. The minimum distance between two successive regions of complete darkness is.....$mm$.

In a $YDSE$,the central bright fringe can be identified:

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

Young's double-slit experiment is performed using green light,red light,and blue light,one at a time. The fringe widths recorded are $\beta_G, \beta_R,$ and $\beta_B$ respectively. Then: