Question

When a thin film of transparent plastic is placed over one of the silts in Young's double silt experiment the central bright firing of the white light firing system is displaced by 4.50 frings. The refractive index of the material is 1.480 and effective wavelength of the light is 550Amstrong. By how much does the film increase the optical path? b. What is the thickness of the film?​

By how much does the film increase the optical path?

A. 2.25 fringes
B. 4.50 fringes
C. 6.75 fringes
D. 9.00 fringes

Answer 1

When a thin film of transparent plastic is placed over one of the slits in Young's double slit experiment, it displaces the central bright fringe. The film increases the optical path by 901.35 Amstrong and has a thickness of 304.46 Amstrong.

Step-by-step explanation:

In Young's double slit experiment, when a thin film of transparent plastic is placed over one of the slits, the central bright fringe of the interference pattern is displaced.

The increase in the optical path can be calculated using the formula: increase in optical path = (fringe displacement * effective wavelength) / (2 * refractive index).

Substituting the given values, we get: increase in optical path = (4.50 fringes * 550 Amstrong) / (2 * 1.480) = 901.35 Amstrong.

Therefore, the film increases the optical path by 901.35 Amstrong.

The thickness of the film can be calculated using the formula: thickness = (increase in optical path) / (2 * refractive index).

Substituting the given values, we get: thickness = (901.35 Amstrong) / (2 * 1.480) = 304.46 Amstrong.

Therefore, the thickness of the film is 304.46 Amstrong.