In an interference experiment the ratio of amplitudes of coherent waves is $${{{a_1}} \over {{a_2}}} = {1 \over 3}$$ . The ratio of maximum and minimum intensities of fringes will be :

In a double-slit experiment, green light (5303$$\mathop A\limits^ \circ $$) falls on a double slit having a separation of 19.44 $$\mu $$m and awidht of 4.05 $$\mu $$m. The number of bright fringes between the first and the second diffraction minima is :

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

Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength $$\lambda $$ such that the first minima occurs directly in front of the slit (S₁) ?