In a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must:

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

In Young's double slit experiment,the slits are horizontal. The intensity at a point $P$ on the screen is $\frac{I_0}{4}$,where $I_0$ is the maximum intensity. If the distance between the two slits $S_1$ and $S_2$ is $d = 2 \lambda$,then the value of $\theta$ is:

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ In Young's double slit experiment,the fringes produced by red light are closer as compared to those produced by blue light.
Reason $(R) :$ The fringe width is directly proportional to the wavelength of light.
In the light of above statements,choose the correct answer from the options given below $:$

If white light is used in Young's double-slit experiment, then a very large number of coloured fringes can be seen. Which of the following statements is correct regarding the fringe pattern?

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 $mm$?