In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

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?

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:

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{57}{2}\right) \lambda$. The band obtained at that point is

In $YDSE$,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width 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.