$A$ double slit interference experiment performed with a light of wavelength $600 \ nm$ forms an interference fringe pattern on a screen with $10^{\text{th}}$ bright fringe having its centre at a distance of $10 \ mm$ from the central maximum. The distance of the centre of the same $10^{\text{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \ nm$ would be . . . . . . $mm$.

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.

Which of the following is a true statement if,in Young's double-slit experiment,the separation between the slits is gradually increased?

The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then:

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:

In a Young's double slit experiment,green light is incident on the two slits. The interference pattern is observed on a screen. Which of the following changes would cause the observed fringes to be more closely spaced?