In Young's double-slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be: (in $: 1$)

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 a Young's double-slit experiment, the angular width of a fringe is $1^\circ$ for light of wavelength $6000 \, \mathring{A}$. What is the distance between the slits in $mm$?

The wavelength of light $500 \, nm$ is used in a Young's double slit experiment. The distance between the slits and the screen is $100 \, cm$ and the slits are separated by $1 \, mm$. Find the distance between the fifth $(5^{th})$ and third $(3^{rd})$ bright fringes. (in $mm$)

Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other.
Reason : The fringe width is inversely proportional to the distance between the two slits.

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$