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:

In the Young's double slit experiment,the distance between the slits varies in time as $d(t) = d_{0} + a_{0} \sin \omega t$; where $d_{0}$,$\omega$,and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

In a Young's double slit experiment,the slit separation $d$ is $0.3 \text{ mm}$ and the screen distance $D$ is $1 \text{ m}$. $A$ parallel beam of light of wavelength $600 \text{ nm}$ is incident on the slits at angle $\alpha$ as shown in the figure. On the screen,the point $O$ is equidistant from the slits and the distance $PO$ is $11.0 \text{ mm}$. Which of the following statement$(s)$ is/are correct?

In Young's double-slit experiment,the slits are separated by $0.2 \, cm$ and illuminated by light of wavelength $\lambda = 5896 \, \mathring{A}$. What is the fringe width on a screen placed $1 \, m$ away from the plane of the slits? What will be the fringe width if the entire system is immersed in water (in $, mm$)? (Refractive index of water $\mu_w = 4/3$)

In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?

In the Young's double slit experiment,if the phase difference between the two waves interfering at a point is $\phi$,the intensity at that point can be expressed by the expression: