In Young's double-slit experiment using monochromatic light,the shape of the interference pattern on the screen is .....

Light consisting of plane waves of wavelengths $\lambda_1 = 8 \times 10^{-5} \ cm$ and $\lambda_2 = 6 \times 10^{-5} \ cm$ generates an interference pattern in Young's double-slit experiment. If $n_1$ denotes the $n_1^{\text{th}}$ dark fringe due to light of wavelength $\lambda_1$ which coincides with the $n_2^{\text{th}}$ bright fringe due to light of wavelength $\lambda_2$, then:

Light of wavelength $520\, nm$ passing through a double slit produces an interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. The separation $d$ between the slits is

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 Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

In $Y.D.S.E$,the ratio of intensity of maxima and minima is $25 : 9$. The ratio of slit width is