White light is passed through a double slit and interference is observed on a screen $1.5 \, m$ away. The separation between the slits is $0.3 \, mm$. The first violet and red fringes are formed $2.0 \, mm$ and $3.5 \, mm$ away from the central white fringes. The difference in wavelengths of red and violet light is $.... \, nm$.

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

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:

In Young's double-slit experiment,the fringe width is $12 \ mm$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$,then the fringe width becomes (in $mm$):