In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

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.

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

Minimum thickness of a mica sheet having $\mu = 3/2$ which should be placed in front of one of the slits in $YDSE$ is required to reduce the intensity at the centre of the screen to half of the maximum intensity is-

In a $YDSE$,light of two different wavelengths $(\lambda_1, \lambda_2)$ is incident normal to the plane of slits. The $n^{th}$ maxima of $\lambda_1$ coincides with the $m^{th}$ maxima of $\lambda_2$ exactly in front of one of the slits. Given $D = 1.5 \ m$,$d = 3 \ mm$,and $4500 \ \mathring{A} < \lambda_1, \lambda_2 < 7000 \ \mathring{A}$,then $n, m$ and $\lambda_1$ are:

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 $........\mu m$.