In Young's double slit experiment, the fringe width is $$12 \mathrm{~mm}$$. If the entire arrangement is placed in water of refractive index $$\frac{4}{3}$$, then the fringe width becomes (in mm):

Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)

Using Young's double slit experiment, a monochromatic light of wavelength 5000 $$\mathop A\limits^o $$ produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000 $$\mathop A\limits^o $$ is used and the separation between the slits is doubled, then the new fringe width will be :

In Young's double slit experiment performed using a monochromatic light of wavelength $$\lambda$$, when a glass plate ($$\mu$$ = 1.5) of thickness x$$\lambda$$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be :