A beam of light having wavelength $$5400 \mathrm{~A}$$ from a distant source falls on a single slit $$0.96 \mathrm{~mm}$$ wide and the resultant diffraction pattern is observed on a screen $$2 \mathrm{~m}$$ away. What is the distance between the first dark fringe on either side of central bright fringe?

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $$\pi / 2$$ at point $$\mathrm{A}$$ and $$\pi$$ at point $$\mathrm{B}$$. Then the difference between the resultant intensities at $$\mathrm{A}$$ and $$\mathrm{B}$$ is

In Young's double slit experiment, the intensity at a point where path difference is $$\frac{\lambda}{6}$$ ($$\lambda$$ being the wavelength of light used) is $$I^{\prime}$$. If '$$I_0$$' denotes the maximum intensity, then $$\frac{I}{I_0}$$ is equal to $$\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{\lambda}\right)$$

In Young's double slit experiment, the distance of $$\mathrm{n}^{\text {th }}$$ dark band from the central bright band in terms of bandwidth '$$\beta$$' is