Question

A laboratory dish, 30 cm in diameter, is half filled with a glucose/water solution. One at a time, 0.45 ?L drops of oil from a micropipette are dropped onto the surface of the solution, where they spread out into a uniform thin film. As more drops are added, the intensity of 630 nm light reflected from the surface decreases. The intensity reaches a minimum after a total of 17 drops have been added. In all questions, assume that the oil's index of refraction is smaller than the glucose/water solution's index.

What is the index of refraction of the oil?

Answer 1

The oil drop can have a refractive index either greater than that of water or less than that of water.

Now, when a ray of light travelling in a medium of low refractive index(low n) reflects from the surface of a medium of high n, there occurs a phase change of 'pi' or half a wavelength path difference is introduced alongwith the reflection too.

Since, the ray going travelling in air and reflecting from oil surface will definitely encounter a 'pi' phase change, the other light wave that travels inside oil will encounter a phase change of 'pi' if the refractive index of water is higher than that of oil and won't encounter a phase change is refractive index of water is lower than oil.

Since as soon as we put 1 drop of oil, there's a very low intensity of light observed due to interference, we can safely say that the refractive index of oil is higher than that of water, since only then will the phase change of 'pi' will be introduced between two reflected waves.

diameter = 30 cm = 0.3 m, radius = 0.15m

Now, 1 oil drop has volume 0.45*10^-9 m³, total height covered by 1 oil drop = 0.45*10^-9/(pi*0.15*0.15) = 6.36*10^-8m

After 17drops, minima occurs,

path difference introduced by 1 oil drop = 2t*n ... where t = 6.36*10^-8m

path difference introduced by 17 drops = 34t*n = lambda = 630*10^-9 ..... n = 0.291