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 wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: If Young's double slit experiment is performed in an optically denser medium than air,then the consecutive fringes come closer.
Reason $(R)$: The speed of light reduces in an optically denser medium than air while its frequency does not change.
In the light of the above statements,choose the most appropriate answer from the options given below:

The ratio of intensities at two points on the screen in Young's double slit experiment when waves from the two slits have a path difference of $0$ and $\lambda/4$ ($\lambda$ is the wavelength of light used) $(\cos 0^\circ = 1, \cos \pi/2 = 0)$.

In Young's double-slit experiment,monochromatic light is used to illuminate the two slits $A$ and $B$. Interference fringes are observed on a screen placed in front of the slits. If a thin glass plate is placed normally in the path of the beam coming from slit $A$,then:

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{57}{2}\right) \lambda$. The band obtained at that point is