List of top Physics Questions on wave interference asked in KEAM

In Young's double slit experiment, monochromatic light of 500 nm falls on the slits separated by a distance of 2 mm. If the screen is 2 m away from the source, then the fringe width is

In a Young's double slit experiment, the intensity at a point where the path difference is \( \frac{\lambda}{6} \) (\( \lambda \) being the wavelength of light used) is \( I \). If \( I_{0} \) denotes the maximum intensity, \( I/I_{0} \) is equal to:

If two waves of equal amplitude \( A \) and opposite phase interfere, the amplitude of the resultant wave is

Light energy is redistributed in:

In Young's double slit experiment using a source of wavelength \(\lambda\) interference bands are observed on a screen. If the separation between the slits alone is halved in this experiment, then the angular separation \(\omega\) of the fringes on the screen becomes

In Young’s double slit experiment performed in air medium, the fringe width observed is 1.4 mm. If the entire arrangement is kept in a liquid medium of refractive index 1.4, then the fringe width (in mm) will be

If two waves of intensities \(I\) and \(4I\) superpose, the ratio between maximum and minimum intensities is

A monochromatic source of wavelength 600 nm was used in Young's double slit experiment to produce interference pattern. $I_1$ is the intensity of light at a point on the screen where the path difference is 150 nm. The intensity of light at a point where the path difference is 200 nm is given by

In Young’s double slit experiment, if the separation between the slits is halved, and the distance between the slits and the screen is doubled, then the fringe width compared to the unchanged one will be

In Young's double slit experiment, to increase the fringe width

In Young's double slit experiment, to increase the fringe width

In Young's double slit experiment, to increase the fringe width