The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

In a Young's double slit experiment,a laser light of $560\,nm$ produces an interference pattern with consecutive bright fringes' separation of $7.2\,mm$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1\,mm$. The wavelength of the second light is $......nm$.

In a Young's double slit experiment,the fringe width is found to be $0.4 \, mm$. If the whole apparatus is immersed in water of refractive index $4/3$ without disturbing the geometrical arrangement,the new fringe width will be

In $\text{YDSE}$,$S_1$ and $S_2$ have the same intensity $I_0$. Column-$I$ shows the distance $x$ of a point $P$ from the central point $O$ on the screen,and Column-$II$ shows the intensity at $P$. Match Column-$I$ with Column-$II$. (Wavelength is $\lambda$)

Column-$I$	Column-$II$
$(A) x = \frac{D \lambda}{d}$	$(P) I_0$
$(B) x = \frac{D \lambda}{4d}$	$(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$	$(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$	$(S) 4 I_0$

In a double-slit experiment,the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed opposite to one of the slits,its order $n$ is:

In Young's double-slit interference experiment,the ratio of the slit widths is $1:25$. Find the ratio of the maximum to minimum intensity in the interference pattern.