In a Young's double-slit experiment,the distance between the two slits is $d = \lambda / 4$,where $\lambda$ is the wavelength of the light used. The initial phase difference is $\pi / 4$. What is the intensity at $\theta = 30^o$?

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

Using Young's double slit experiment,a monochromatic light of wavelength $5000 \,\mathring A$ produces fringes of fringe width $0.5 \,mm$. If another monochromatic light of wavelength $6000 \,\mathring A$ is used and the separation between the slits is doubled,then the new fringe width will be ............... $mm$.

The figure shows a schematic diagram of the arrangement of Young's Double Slit Experiment. If the distance $d$ is varied,then identify the correct statement.

In a Young's double-slit experiment with light of wavelength $\lambda$,the separation of slits is $d$ and the distance of the screen is $D$ such that $D >> d >> \lambda$. If the fringe width is $\beta$,the distance from the point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

The ratio of intensities at two points $P$ and $Q$ on the screen in a Young's double-slit experiment,where the phase differences between two waves of the same amplitude are $\pi / 3$ and $\pi / 2$ respectively,is: