In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is _______.
In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm, the angular width of central maximum is $\alpha \times 10^{-2}$ degrees. The value of $\alpha$ is ________.
A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ________ $\times 10^{-5}$ m.
In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are $ 2 : 1 $ and $1: 2$ respectively. The corresponding ratio of the distances between the slits and the respective screens ( $D_1 / D_2$ ) is $\_\_\_\_$。
JEE Main Subjects
Browse all chapters by subject