In Young's double slit interference experiment,the slit separation is made $3$ fold. The fringe width becomes

In a Young's double-slit experiment,the fringe width is $0.2 \, mm$. If the wavelength of light is increased by $10\%$ and the distance between the slits is increased by $10\%$,then the new fringe width will be ..... $mm$.

In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

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

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .