A thin 1-m long rod has a 200-g object attached to one end and a 300-g object attached to the
other end. At what point should the rod be supported so that it remains in horizontal
equilibrium?
(A) 20 cm from the 300-g object
(B) 40 cm from the 300-g object
(C) 60 сm from the 300-g object
(D) 40 cm from the 200-g object

a)  312.5 10⁻⁹ m    and b)  t = 156 10⁻⁹ m

Explanation:

When analyzing your problem we see that it is a phenomenon of interference of thin films.

a) In this case the girl is in the air that has a refractive index of n = 1, so the light beam when it leaves the film with a higher refractive index has a 180º phase change. When the beam advances inside the film it finds another medium with water with refractive index n3 = 1.33, as this is greater than that of oil (n3> n2) undergoes another 180 ° phase change, so we can see that the film experiences in total two phase change.

In addition, this experiment is performed almost perpendicularly, so the sine tes = 1, therefore, the constructive interference ratio is

    2 t = (m) λₙ

Where t is the thickness of the film, m is an integer and, λₙ is the wavelength that comes out of the film.

When the girl throws the beam from the spectrometer, it has a wavelength of 750 nm, it reaches the film and its wavelength changes with the refractive index of the film, the frequency of the wave remains constant.

    λₙ= λ₀ / n2

Where λ₀ is the wavelength of the air and the refractive index of the film

Afora we can substitute and calculate the thickness (t)

    t = ½ (m) λ₀ / n

The minimum thickness occurs for m = 1

    t = ½ (1) λ₀ / n

    t =  ½  750 10-9 / 1.20

    t = 312.5 10⁻⁹ m

b) in this case when it reaches the second medium (n3 <n2) it does not undergo any phase change, so the constructive interference equation that

    2t = (m + ½) λₙ

    2t = (m + ½) λ₀ / n

The ½ well of the phase change when passing from air to oil

For a minimum thickness m = 0

    t = ½ (0+ ½) 750 / 1.20

    t = ¼ 750/ 1.20

    t = 156 10⁻⁹ m