A thin film of soap solution is illuminated by white light. In the reflected

Question

light, two dark consecutive overlapping fringes are observed
corresponding to wavelengths 6.1 x 10^-7 m and 6.0 x 10^
-7 m. The
refractive index for soap solution is 4/3. Calculate the thickness of the
film.

Answer 1

To solve this problem, we'll use the formula for the thickness of a thin film:

t = (m + 1/2) * λ / (2 * n)

where:
t is the thickness of the film
m is the order number of the fringe
λ is the wavelength of light in the film
n is the refractive index of the film

First, let's calculate the difference in path lengths (∆d) between the two consecutive fringes that we observed.

∆d = (m + 1) * λ_1 - m * λ_2

where:
λ_1 = 6.1 x 10^-7 m is the wavelength of the first fringe
λ_2 = 6.0 x 10^-7 m is the wavelength of the second fringe

∆d = (1 + 1) * (6.1 x 10^-7) - 1 * (6.0 x 10^-7)
= 1.2 x 10^-7 m

Now let's calculate the average wavelength of light in the film, λ_avg:

λ_avg = (λ_1 + λ_2) / 2
= (6.1 x 10^-7 + 6.0 x 10^-7) / 2
= 6.05 x 10^-7 m

Finally, we can substitute these values into the formula to find the thickness of the film:

t = (∆d * λ_avg) / (2 * n)
= (1.2 x 10^-7 * 6.05 x 10^-7) / (2 * 4/3)
= (7.26 x 10^-14) / (8/3)
= (7.26 x 10^-14) * (3/8)
= 2.7375 x 10^-14 m
= 2.74 x 10^-14 m (rounded to two significant figures)

Therefore, the thickness of the film is 2.74 x 10^-14 m.