Minimum thickness of a mica sheet having $\mu = 3/2$ which should be placed in front of one of the slits in $YDSE$ is required to reduce the intensity at the centre of the screen to half of the maximum intensity is-

In a Young's double-slit experiment,the distance between the two slits is twice the wavelength of the light used. Find the total number of bright fringes formed on the screen.

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 an interference pattern,the fringe width is $\beta$. If the frequency of the source is doubled,the fringe width becomes .....

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

In Young's double slit experiment, the angular width of a fringe is found to be $0.2^\circ$ on a screen placed $1 \text{ m}$ away. The wavelength of light used is $600 \text{ nm}$. If the entire apparatus is immersed in water of refractive index $4/3$, the angular width of the fringe will be: (in $^\circ$)