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 a Young's double-slit experiment,light of wavelength $5890 \ \mathring A$ is used,and the angular fringe width on the screen is $0.20^\circ$. If the entire apparatus is immersed in water,find the new angular fringe width. (Refractive index of water $\mu = 4/3$) (in $^\circ$)

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and are illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the plane of the slits and the source slit is $D$. $A$ screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is

In Young's double-slit experiment,the distance between the two slits is $0.2 \, mm$ and the screen is $200 \, cm$ away from the slits. If the wavelength of light used is $5000 \, \mathring{A}$,find the distance of the third bright fringe from the central bright fringe in $cm$.

In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

In Young's double slit experiment,the slits are horizontal. The intensity at a point $P$ on the screen is $\frac{I_0}{4}$,where $I_0$ is the maximum intensity. If the distance between the two slits $S_1$ and $S_2$ is $d = 2 \lambda$,then the value of $\theta$ is: