In a $Young's$ double-slit experiment,light of wavelength $6000 \, \mathring{A}$ is used. If the third dark fringe is formed at point $P$ on the screen,then the path difference $(S_1P - S_2P)$ is ......... $\mu m$.

Two coherent sources of light are placed at points $(-\frac{5a}{2}, 0)$ and $(+\frac{5a}{2}, 0)$. The wavelength of the light is $\lambda = \frac{4a}{3}$. How many maxima will be obtained on a planar circle of large radius with its center at the origin?

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:

Assertion: $A$ white source of light during interference forms only white and black fringes.
Reason: Width of fringe is inversely proportional to the wavelength of the light used.

In Young's double-slit experiment,the distance between the two slits is $0.2 \, mm$ and the screen is $200 \, cm$ away from the slits. If the wavelength of light used is $5000 \, \mathring{A}$,find the distance of the third bright fringe from the central bright fringe in $cm$.

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: