In the Young’s double slit experiment using a monochromatic light of wavelength λ , the path
difference ( in terms of an integer n) corresponding to any point having half the peak
intensity is

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π/3, 2π/3 and π . When they are superposed, the intensity of the resulting wave is nI0. The value of n is

In Young’s double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference Φ is given by :

Two coherent sources of light can be obtained by :

Two identical light waves, propagating in the same direction, have a phase difference δ. After they superpose, the intensity of the resulting wave will be proportional to :

If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima is :

In the Young’s double slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that :-
(A) The intensities of individual sources are 5 and 4 units respectively
(B) The intensities of individual sources are 4 and 1 units respectively
(C) The ratio of their amplitudes is 3
(D) The ratio of their amplitudes is 2

In a Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3 without disturbing the geometrical arrangement, the new fringe width will be :

In a Young’s double slit experiment the intensity at a point where the path difference is λ/6 (λ being the wavelength of the light used) is I. If I0 denotes the maximum intensity, I/I0 is equal to :

In the Young’s double slit experiment, for which colour the fringe width is least :