In Young's double-slit experiment,if one slit is covered with a blue filter and the other with a yellow filter,then the interference pattern will be:

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

In a double slit experiment,when the distance between slits is increased $10$ times,while their distance from the screen is halved,then the fringe width

For Young's double slit experiment,two statements are given below :
Statement $I:$ If the screen is moved away from the plane of the slits,the angular separation of the fringes remains constant.
Statement $II:$ If the monochromatic source is replaced by another monochromatic source of larger wavelength,the angular separation of the fringes decreases.
In the light of the above statements,choose the correct answer from the options given below :

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

In Young's double slit experiment,the angular width of fringes is $0.20^o$ for sodium light of wavelength $5890 \ \mathring A$. If the complete system is dipped in water,then the angular width of the fringes becomes....$^o$