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 a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

In a Young's double-slit experiment, interference fringes are obtained on a screen at a distance of $1 \, m$ using light of wavelength $6000 \, \mathring{A}$. The distance between the slits is $1 \, mm$. The fringe width is:

In a Young's double-slit experiment,the distance between the two sources is $0.1 / \pi \, mm$. The distance between the source and the screen is $25 \, cm$. The wavelength of light is $5000 \, \mathring{A}$. The angular position of the first dark fringe is ........$^o$.

In a double-slit experiment performed in air, the angular width of a fringe is found to be $0.15^{\circ}$ on a screen placed $80 \ cm$ away. The wavelength of light used is $490 \ nm$. What is the angular width of the fringe if the entire apparatus is immersed in a medium of refractive index $\frac{5}{3}$ (in $^{\circ}$)?

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$