Question

A student is asked to determine the index of refraction of a glass slab. She conducts several trials for measurement of angle of incidence θ_a in the air versus angle of refraction θ_g in the glass at the surface of the slab. She records her data in the following table. The index of refraction in air is 1.0.

(a) Plot the data points on the axes below and draw a best-fit line through the points.

(b) Calculate the index of refraction of the glass slab from your best-fit line.

(c) Describe how you could use the graph to determine the critical angle for the glass-air interface. Do not use the answer to the part (b) for this purpose.

(d) On the graph in (a), sketch and label a line for a material of higher index of refraction.

Answer 1

(a) a is the lower slope line

(b) From Snells law, the slope of this graph is n_glass = 1.5

(c) From n_glass sin θ_c = n_air sin θ_air … we want to use the point where sin θ_air = sin 90 = 1. This point will correspond to the related angle in the glass and since it’s the critical point, this will allow you to find θ_c

So we find sin θ_a = 1 from the graph and find the corresponding value of sin θ_glass that goes with it.

Extending the lower slope line up to 1.0, we get a sin θ_g value and then set that value = sin θ_g and solve for θ which is the critical angle.

(d)