In a Young double slit experiment, the wavelength of incident light is $6000 \mathop {\rm{A}}\limits^{\rm{o}}$, the separation between slits $S_1$ and $S_2$ is 5 cm and the distance between slits plane and screen is 50 cm , as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is $\_\_\_\_$ Å.

In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda / 3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda / 6$. The ratio of intensities at points $A$ and $B$ is $\_\_\_\_$ .

The maximum intensity in a Young's double slit experiment is $I_0$. Distance between the slits $(d)$ is $5 \lambda$, where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D=10 d$ is $\_\_\_\_$ .

In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4 \mu \mathrm{~m}$. If the experiment is carried out in another medium having refractive index 1.2 , the fringe width will be $\_\_\_\_$ $\mu \mathrm{m}$.