In a Young's double slit experiment,the slit separation is $1 \, mm$ and the screen is $1 \, m$ from the slit. For a monochromatic light of wavelength $500 \, nm$,the distance of the $3rd$ minima from the central maxima is.....$mm$

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement-$1$: In Young's double slit experiment,the number of fringes observed in the field of view is small with longer wavelength of light and is large with shorter wavelength of light.
Statement-$2$: In the double slit experiment,the fringe width depends directly on the wavelength of light.

Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:

In Young's double-slit experiment,the position of the $5^{\text{th}}$ bright fringe from the central maximum is $5\,cm$. The distance between the slits and the screen is $1\,m$ and the wavelength of the monochromatic light used is $600\,nm$. The separation between the slits is $........\mu m$.

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

Two slits separated by a distance of $1\, mm$ are illuminated with red light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed $1\, m$ from the slits. Find the distance between the third dark fringe and the fifth bright fringe on the same side of the central maxima.