In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$

In an interference arrangement similar to Young's double slit experiment,the slits $S_1$ and $S_2$ are illuminated with coherent microwave sources each of frequency $10^6 \ Hz$. The sources are synchronized to have zero phase difference. The slits are separated by distance $d = 150 \ m$. The intensity $I(\theta)$ is measured as a function of $\theta$,where $\theta$ is defined as shown. If $I_0$ is the maximum intensity,then $I(\theta)$ for $0 \le \theta \le 90^\circ$ is given by:

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width,the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is ..... .

Young's double slit experiment is performed with light of wavelength $550 \, nm$. The separation between the slits is $1.10 \, mm$ and the screen is placed at a distance of $1 \, m$. What is the distance between consecutive bright or dark fringes in $mm$?

In the Young's double slit experiment,if the phase difference between the two waves interfering at a point is $\phi$,the intensity at that point can be expressed by the expression: