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:

Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other.
Reason : The fringe width is inversely proportional to the distance between the two slits.

In a Young's double-slit experiment,the slits are $2 \, mm$ apart and are illuminated by a mixture of two wavelengths $\lambda_1 = 7500 \, \mathring{A}$ and $\lambda_2 = 9000 \, \mathring{A}$. At what distance from the central maximum on a screen $2 \, m$ away will a bright fringe from one interference pattern coincide with a bright fringe from the other?

In a Young's double slit experiment,the angular width of a fringe is $0.35^{\circ}$ on a screen placed at $2\,m$ away for a particular wavelength of $450\,nm$. The angular width of the fringe,when the whole system is immersed in a medium of refractive index $7/5$,is $\frac{1}{\alpha}$. The value of $\alpha$ is ..............

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

In Young's double-slit experiment,the distance between the two slits is $2 \times 10^{-3} \, m$ and the distance between the slits and the screen is $2.5 \, m$. The wavelength of the light used ranges from $2000 \, \mathring{A}$ to $9000 \, \mathring{A}$. What wavelength (in $\mathring{A}$) will form a bright fringe at a distance of $10^{-3} \, m$ from the central maximum?