The thickness of a plate having refractive index $\mu$ which will produce a change in optical path equal to half the wavelength $\lambda$ of the light passing through it is

The following diagram is an arrangement to view the interference pattern produced by a wedge of liquid between a microscope slide and a glass block. The interference pattern obtained is made of equally spaced parallel fringes. The fringe separation (or fringe width) $\beta$ may be decreased by some of the following actions (Consider one change at a time).

When a mica sheet of thickness $7 \ \mu m$ and refractive index $\mu = 1.6$ is placed in the path of one of the interfering beams in a biprism experiment,the central fringe shifts to the position of the $7^{th}$ bright fringe. Find the wavelength of the light used.

$A$ thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate with the curved surface downwards. Monochromatic light is incident normally from the top. The observed interference fringes from this combination do not follow one of the following statements.

In a biprism experiment, the fifth dark fringe is obtained at a point. $A$ thin transparent film of refractive index '$\mu$' is placed in one of the interfering paths. Now, the $7^{th}$ bright fringe is obtained at the same point. If '$\lambda$' is the wavelength of light used, the thickness of the film is equal to:

In a biprism experiment,the $6^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $7^{\text{th}}$ dark band with wavelength $\lambda_2$. Then the ratio $\lambda_1 : \lambda_2$ is (other settings remain the same):