Interference fringes and their formation

Interference fringes are formed when two or more coherent waves superpose and interfere with each other. In the case of interference fringes formed by the superposition of two waves, such as in Young's double-slit experiment or in a Michelson interferometer, the resulting pattern consists of bright and dark regions called fringes.

Wavelength and number of fringes

The wavelength of light plays a crucial role in the formation of interference fringes. As light passes through a medium, its wavelength may change due to the refractive index of the medium. The number of fringes that are formed depends on the path difference between the interfering waves.

Shift in central fringe with a plate of material

When a plate of material is placed in the path of one of the interfering beams, it causes a shift in the position of the interference fringes. This shift occurs because the plate introduces an additional path difference between the two beams.

In this case, the central fringe is shifted to a position occupied by the 10th fringe. This means that the path difference introduced by the plate is equal to the path difference between the central fringe and the 10th fringe.

Calculating the path difference

To calculate the path difference introduced by the plate, we can use the formula:

Path difference = t * (refractive index - 1)

where t is the thickness of the plate and refractive index is the index of refraction of the material.

Since the central fringe is shifted to the position of the 10th fringe, the path difference is equal to the path difference between the central fringe and the 10th fringe, which can be calculated using the formula:

Path difference = (10 * wavelength) / 2

where wavelength is the wavelength of the light used.

Solving for the refractive index

By equating the two path differences, we can solve for the refractive index of the material:

t * (refractive index - 1) = (10 * wavelength) / 2

Simplifying the equation, we get:

(refractive index - 1) = (10 * wavelength) / (2 * t)

refractive index = ((10 * wavelength) / (2 * t)) + 1

Substituting the given values of wavelength (600 nm) and t (0.001 cm), we can calculate the refractive index of the material.

Visually, the interference fringes are formed when two coherent waves superpose and interfere with each other. These fringes consist of bright and dark regions and their formation depends on the wavelength of light and the path difference between the interfering waves. When a plate of material is introduced in the path of one beam, it causes a shift in the position of the interference fringes. In this case, the central fringe is shifted to the position of the 10th fringe. By calculating the path difference introduced by the plate and equating it to the path difference between the central fringe and the 10th fringe, we can solve for the refractive index of the material.

1.6