In Young's double-slit experiment,an interference pattern is obtained on a screen by light of wavelength $6000 \ \mathring A$,coming from coherent sources $S_1$ and $S_2$. At a certain point $P$ on the screen,the third dark fringe is formed. Then the path difference $S_1P - S_2P$ in microns is:

$A$ beam of light consisting of two wavelengths, $650\; nm$ and $520\; nm$, is used to obtain interference fringes in a Young's double-slit experiment.
$(a)$ Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650\; nm$.
$(b)$ What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Consider the following statements in the case of Young's double-slit experiment:
$(1)$ $A$ slit is necessary if we use an ordinary extended source of light.
$(2)$ $A$ slit is not necessary if we use an ordinary but well-collimated beam of light.
$(3)$ $A$ slit is not needed if we use a spatially coherent point source of light.
Which of the above statements is true?

In a double slit experiment shown in the figure,when light of wavelength $400 \ nm$ is used,a dark fringe is observed at $P$. If $D=0.2 \ m$,the minimum distance between the slits $S_1$ and $S_2$ is . . . . . . $mm$.

In Young's double slit experiment,the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same setup,the two slits are sources of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is:

Two slits,$4 \, mm$ apart,are illuminated by light of wavelength $6000 \, \mathring{A}$. What will be the fringe width on a screen placed $2 \, m$ from the slits in $mm$?