In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is :

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 $\_\_\_\_$ .