To increase fringe width, by keeping distance between slit and screen constant, we need to ensure that,

In Young's double slit experiment, If a third slit is made in between the double slits then?

Two coherent light sources $S_1$ and $S_2$ $(\lambda =6000\mathit{ }\mathrm{\AA })$ are $1 \mathrm{mm}$ apart from each other. The screen is placed at a distance of $25 \mathrm{cm}$ from the sources. The width of the fringes on the screen should be,

In a certain double slit experimental arrangement, interference fringes of width $1.0 \mathrm{mm}$ each are observed when light of wavelength $5000 \mathrm{\AA }$ is used. Keeping the set-up unaltered, if the source is replaced by another source of wavelength $6000 \mathrm{\AA }$, the fringe width will be,

In a Young's double slit experiment, the fringe width is found to be $0.4 \mathrm{mm}$. If the whole apparatus is immersed in water of refractive index $\frac{4}{3}$ without disturbing the geometrical arrangement, the new fringe width will be,

In Young's double slit experiment, if $L$ is the distance between the slits and the screen upon which interference pattern is observed, $X$ is the average distance between the adjacent fringes and $d$ being the slit separation, the wavelength of light is given by,

In Young's double slit experiment carried out with light of wavelength $\lambda =5000 \mathrm{\AA }$, the distance between the slits is $0.2 \mathrm{mm}$ and the screen is at $200 \mathrm{cm}$ from the slits. The central maximum is at $x=0$. The third maximum (taking the central maximum as zero^th maximum) will be at $x$ equal to

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by $4\times {10}^{-2} \mathrm{m}$ towards the slits, the change in fringe width is $2\times {10}^{-5} \mathrm{m}$. If the separation between the slits is ${10}^{-3} \mathrm{m}$, then the wavelength of light used is

In a double slit interference experiment, the distance between the slits is $0.05 \mathrm{cm}$ and screen is $2 \mathrm{m}$ away from the slits. The wavelength of light is $8.0\times {10}^{-5} \mathrm{cm}$. The distance between successive fringes is