Ascientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). she aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). the index of refraction of water is 1.33.

(a) the index of refraction of the oil is 1.20. what is the minimum thickness t of the oil slick at that spot?
(b) suppose the oil had an index of refraction of 1.50. what would the minimum thickness t be now?

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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards