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

The angular width of a fringe in a double slit experiment is found to be $0.2^{\circ}$ on a screen $1 \ m$ away. The wavelength of light used is $600 \ nm$. What is the change in the angular width of the fringe if the entire measurement system is immersed in water (in $^{\circ}$)? (Use refractive index of water as $\mu = \frac{4}{3}$)

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

$A$ 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:

On using red light $(\lambda = 6600 \ \text{Å})$ in Young's double slit experiment, $60$ fringes are observed in the field of view. If violet light $(\lambda = 4400 \ \text{Å})$ is used, the number of fringes observed will be

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is: