Young's double slit experiment is carried out by using green, red and blue light; one colour at a time. The fringe widths recorded are $\beta_{\mathrm{G}}, \beta_{\mathrm{R}}$ and $\beta_{\mathrm{B}}$, respectively. Then 

In Young's double slit experiment, if wavelength of light is doubled without changing other conditions, the fringe width will

In Young's double-slit experiment with a source of light of wavelength $6320 \stackrel{\circ }{\mathrm{A}}$ the first maxima will occur when

Bandwidth for the red light of wavelength $6400 \stackrel{\circ }{\mathrm{A}}$ is $0.32 \mathrm{mm}$. If the red light is replaced by the blue light of wavelength $4800 \stackrel{\circ }{\mathrm{A}}$, then the change in bandwidth will be 

In Young's double-slit experiment, when the wavelength used is $6000 \stackrel{\circ }{\mathrm{A}}$ and the screen is $40 \mathrm{cm}$ from the slits, the fringes are $0.012 \mathrm{cm}$ wide. What is the distance between the slits?

In a double-slit experiment, the distance between the slit is $1 \mathrm{mm}$ and the screen is $25 \mathrm{cm}$ away from the slits. The wavelength of light is $6000 \stackrel{\circ }{\mathrm{A}}$. The width of the fringe on the screen is

A monochromatic light beam of wavelength $5896 \stackrel{\circ }{\mathrm{A}}$ is used in double-slit experiment to get an interference pattern on a screen. $9^{\mathrm{th}}$ bright fringe is seen at a particular position on the screen. At the same point on the screen, if ${11}^{\mathrm{th}}$ bright fringe is to be seen, the wavelength of the light that is needed is (nearly)

The path difference at a point on the screen in Young's experiment is $5\lambda$. If the distance of that point from the central bright band is $0.5 \mathrm{mm}$, then the bandwidth is 

In Young's experiment, the distance between the slits is reduced to half and the distance between the slit and screen is doubled, then the fringe width