Question

For a light of wavelength 6 × 10 ^7 m it is found that in a thin film of air, 9 fringes occur between two points. Deduce the difference of film thickness between these points​

Answer 1

Explanation:

State two differences between interference and diffraction For light of wavelength λ = 6 × 10–7 m, it is found that in a thin film of air, 9 fringes occur between two points deduce the difference of film thickness between these points.

plz Mark me Brainlist

Answer 2

Answer:

The difference in film thickness between the two points is 2.7 × 10^8 meters.

Explanation:

The path difference between two interfering waves that create adjacent bright or dark fringes in a thin film is given by the formula:

Δ = 2nt

where Δ is the path difference, n is the refractive index of the film, and t is the thickness of the film.

Since the film is air, we can assume n = 1.

We are given that 9 fringes occur between two points for light of wavelength 6 × 10^7 m. This means that the path difference between these two points is 9 times the wavelength:

Δ = 9λ = 9(6 × 10^7 m) = 5.4 × 10^8 m

Substituting this into the formula, we get:

5.4 × 10^8 m = 2nt

Since n = 1 for air, we can simplify this to:

5.4 × 10^8 m = 2t

Dividing both sides by 2, we get:

t = 2.7 × 10^8 m

Therefore, the difference in film thickness between the two points is 2.7 × 10^8 meters.

To learn more about similar questions visit:

https://brainly.in/question/32753541?referrer=searchResults

https://brainly.in/question/5644469?referrer=searchResults

#SPJ3