A thin air film is between two points A and B. When light of wavelength `5400 Å` is used, 5 fringes appear between the two points. Find the difference in the thickness of the film between the two points.

In a thin air film 8 fringes occurs between two points. The light is incident normally on the film. The wavelength of light is 500 nm. The difference in thickness of the film between two points is

If A and B are the two points. Find the distance between them : .

Using light of wavelength 6,000 overset@A ,it is found that in a thin film of air,7 fringes occur between two points,deduce the difference of the film thickness between these points.

In a wedge-shaped film, between two points X and Y, nine friges are observed with light of wavelength 4800 Å . How many fringes will be observed between the same two points if the wavelength of light is changed to 6400 Å ?

What is the distance between two points H and C ?And find the distance between two points A and B (AS_(1))

The minimum thickness of air film which appears dark under green light of wavelength 500 nm is:

The path difference between two wavefronts emitted by coherent sources of wavelength 5460 Å is 2.1 micron . The phase difference between the wavefronts at that point is –

The value of numerical aperature of the objective lens of a microscope is 1.25 if light of wavelength 5000 Å is used the minimum separation between two points to be seen as distinct, will be