In a $YDSE$ with two identical slits,when the upper slit is covered with a thin,perfectly transparent sheet of mica,the intensity at the centre of the screen reduces to $75\%$ of the initial value. The second minimum is observed to be above this point and the third maximum below it. Which of the following cannot be a possible value of the phase difference caused by the mica sheet?

In Young's double slit experiment,the distance between slits and the screen is $1.0\,m$ and monochromatic light of $600\,nm$ is being used. $A$ person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied,the interference pattern disappears for a particular distance $d_0$ between the slits. If the angular resolution of the eye is $\frac{1}{60}^o,$ the value of $d_0$ is close to......$mm$

In an interference arrangement similar to Young's double slit experiment,the slits $S_1$ and $S_2$ are illuminated with coherent microwave sources each of frequency $10^6 \ Hz$. The sources are synchronized to have zero phase difference. The slits are separated by distance $d = 150 \ m$. The intensity $I(\theta)$ is measured as a function of $\theta$,where $\theta$ is defined as shown. If $I_0$ is the maximum intensity,then $I(\theta)$ for $0 \le \theta \le 90^\circ$ is given by:

In Young's double-slit experiment,the fringe width is $12 \ mm$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$,then the fringe width becomes (in $mm$):

In Young's double slit experiment using sodium light $(\lambda_1 = 5898 \ \text{\AA})$, $92$ fringes are seen. If a light of wavelength $\lambda_2 = 5461 \ \text{\AA}$ is used instead, how many fringes will be seen in the same field of view?

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves: