Question

A student performs an experiment to determine the index of refraction n of a rectangular glass slab in air. She is asked to use a laser beam to measure angles of incidence θi in air and corresponding angles of refraction θr in glass. The measurements of the angles for five trials are given in the table below.

(a) Complete the last two columns in the table by calculating the quantities that need to be graphed to provide a linear relationship from which the index of refraction can be determined. Label the top of each column.

(b) On the grid, plot the quantities calculated in (a) and draw an appropriate graph from which the index of refraction can be determined. Label the axes.

(c) Using the graph, calculate the index of refraction of the glass slab.

The student is also asked to determine the thickness of a film of oil (n = 1.43) on the surface of water (n = 1.33).

Light from a variable wavelength source is incident vertically onto the oil film as shown above. The student measures a maximum in the intensity of the reflected light when the incident light has a wavelength of 600 nm.

(d) At which of the two interfaces does the light undergo a 180 phase change on reflection?

The air-oil interface only The oil-water interface only Both interfaces Neither interface

(e) Calculate the minimum possible thickness of the oil film.

Answer 1

(a)

(b)

(c) From snells law. ni sin θi = nr sin θr … nair sin θi = nglass sin θr … nglass = sin θi / sin θr

For the graph we have, this would be the inverse slope → 1 / slope. Slope = 0.67 … 1/0.67 … nglass = 1.5

(d) λ phase shifts happen when entering a more optically dense material. So only air–oil does a phase shift happen.

(e) First we need to know the wavelength in the film (oil) … n1 λ1 = n2 λ2 … (1)(600) = (1.43)λoil … λoil = 420 nm

Since there is already a λ phase shift from the boundary flip, we need a total extra phase shift of λ from traveling in the film to produce constructive interference (maximum). For this, λoil is required = 105 nm.