In a Young's double slit experiment, the slits are separated by 0.5 mm . Fringes are obtained on a screen which is placed at distance 1 m away from the slits. When the screen is moved 7 cm farther away, the fringe width changes by $63 \mu \mathrm{~m}$. The wavelength of light used in the experiment will be:

The interference pattern is obtained with two coherent light sources of intensity ratio $9: 1$. The ratio of $\frac{I_{\operatorname{MAX}}+I_{\operatorname{MIN}}}{I_{\operatorname{MAX}}-I_{\operatorname{MIN}}}$ is $\frac{\alpha}{\beta}$. The values of $\alpha$ and $\beta$ are:

The wavelength of a monochromatic light which is used in single slit diffraction is 800 nm . The width of the single slit for which the first minimum appears at $\theta=45^{\circ}$ on the screen will be:

Young's double slit experiment is first done in air and then in a medium of refractive index $\mu$. If the 7 th dark fringe in the medium lies where the 4 th bright fringe is in air, then the value of $\mu$ is :