When a thin metal plate is placed in the path of one of the interfering beams of light in a Young's double-slit experiment,what happens to the interference pattern?

In Young's double-slit experiment,both slits are illuminated by a laser beam and the interference pattern was observed on a screen. If the viewing screen is moved farther from the slit,what happens to the interference pattern?

In Young's double slit experiment,the intensity on the screen at a point where path difference is $\lambda$ is $K.$ What will be the intensity at the point where path difference is $\lambda /4$?

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that

In a Young's double slit experiment,the angular width of a fringe is $0.35^{\circ}$ on a screen placed at $2\,m$ away for a particular wavelength of $450\,nm$. The angular width of the fringe,when the whole system is immersed in a medium of refractive index $7/5$,is $\frac{1}{\alpha}$. The value of $\alpha$ is ..............

In Young's double slit experiment,carried out with light of wavelength $5000 \ \mathring{A}$,the distance between the slits is $0.3 \ \text{mm}$ and the screen is at $200 \ \text{cm}$ from the slits. The central maximum is at $x=0 \ \text{cm}$. The value of $x$ for the third maxima is ............. $\text{mm}$.