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?

Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)

Assertion: In Young's double slit experiment,if the wavelength of incident monochromatic light is doubled,the number of bright fringes on the screen will increase.
Reason: The maximum number of bright fringes on the screen is directly proportional to the wavelength of light used.

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

If the separation between slits in Young's double slit experiment is reduced to $\frac{1}{3}rd,$ the fringe width becomes $n$ times. The value of $n$ is

If yellow light emitted by a sodium lamp in Young's double-slit experiment is replaced by monochromatic blue light of the same intensity,then: