A laboratory dish, 20 cm in diameter, is half filled with water. One at a time, 0.50/µL drops of oil from a micropipette are dropped onto the surface of the water, where they spread out into a uniform thin film. After the first drop is added, the intensity of 600 nm light reflected from the surface is very low. As more drops are added, the reflected intensity increases, then decreases again to a minimum after a total of 13 drops have been added. What is the index of refraction of the oil?

n = 1.45

Explanation:

Given:

- Diameter of the dish d = 0.2 m

- Volume per drop V/drop = 0.5 uL

- wavelength of incident light λ = 600 nm

- minimum intensity @ 13 drops

- refractive index of oil : n

Find:

What is the index of refraction of the oil?

Solution:

- Destructive interference can be seen @ 13 drops. So we will use the relation of destructive interference:

n = m*λ / 2*t

- Compute the thickness of the film.

Volume of film = 0.5*13*10^-9 = 6.5*10^-9 m^3

Volume of cylinder = pi*d^2 / 4 * t

pi*d^2 / 4 * t = 6.5*10^-9 m^3

t = 4 * (6.5*10^-9) / pi*0.2^2 = 2.069*10^-7 m

- Use the given relation @ m = 1:

n = λ / 2*t

n = (600*10^-9) / 2*2.069*10^-7

n = 1.45