Two coherent sources of light interfere. The intensity ratio of two sources is $$1: 4$$. For this interference pattern if the value of $$\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$$ is equal to $$\frac{2 \alpha+1}{\beta+3}$$, then $$\frac{\alpha}{\beta}$$ will be :

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 :