In $YDSE$,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

The distance between two slits in a double slit experiment is $1 \ mm$. The distance between the slits and the screen is $1 \ m$. If the distance of the $10^{th}$ fringe from the central fringe is $5 \ mm$, then the wavelength of light is $... \mathring{A}$.

If yellow light in the Young's double slit experiment is replaced by red light,the fringe width will

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:

In Young's double slit experiment, the distance between the two slits is $0.1 \, mm$ and the wavelength of light used is $4 \times 10^{-7} \, m$. If the width of the fringe on the screen is $4 \, mm$, the distance between the screen and the slit is:

In an interference arrangement similar to Young's double slit experiment,the slits $S_1$ and $S_2$ are illuminated with coherent microwave sources each of frequency $10^6 \ Hz$. The sources are synchronized to have zero phase difference. The slits are separated by distance $d = 150 \ m$. The intensity $I(\theta)$ is measured as a function of $\theta$,where $\theta$ is defined as shown. If $I_0$ is the maximum intensity,then $I(\theta)$ for $0 \le \theta \le 90^\circ$ is given by: