Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:

$A$ beam of light consisting of two wavelengths $650 \, nm$ and $520 \, nm$ is used to illuminate the slits of a Young's double-slit experiment. The order of the bright fringe of the longer wavelength that coincides with a bright fringe of the shorter wavelength at the least distance from the central maximum 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 a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

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:

In Young's double slit experiment using sodium light $(\lambda_1 = 5898 \ \text{\AA})$, $92$ fringes are seen. If a light of wavelength $\lambda_2 = 5461 \ \text{\AA}$ is used instead, how many fringes will be seen in the same field of view?