Use the diagram above as a guide to carry out the following experiment. Trace the outline ABCD of the rectangular glass prism on the drawing paper provided....

Answer Details

(a) Refraction of light through a rectangular glass block

The ray RN strikes face AB at N, refracts into the glass and travels to O on face DC. LNM is the normal at N and M is the foot of the normal on DC, so NM is perpendicular to DC and triangle NMO is right-angled at M.

In right-angled triangle NMO the refracted ray NO makes the angle of refraction \( r \) with the normal NM, so \( \sin r = \dfrac{MO}{NO} \). The glancing angle at AB is \( \theta \), hence the angle of incidence is \( i = 90^\circ - \theta \) and \( \sin i = \cos\theta \). Applying Snell's law \( \sin i = n\sin r \):

\[ \cos\theta = n\,\frac{MO}{NO} \]

A graph of \( \cos\theta \) against \( \varphi = \dfrac{MO}{NO} \) is therefore a straight line through the origin whose slope is the refractive index \( n \).

Observation / table of readings

S/N	θ (°)	MO (cm)	NO (cm)	φ = MO/NO	cos θ
1	75.0	1.1	6.1	0.18	0.26
2	65.0	1.8	6.3	0.29	0.42
3	55.0	2.5	6.5	0.38	0.57
4	45.0	3.2	6.8	0.47	0.71
5	35.0	3.9	7.2	0.54	0.82

Worked check for row 1: \( \varphi = \dfrac{MO}{NO} = \dfrac{1.1}{6.1} = 0.18 \) and \( \cos 75^\circ = 0.26 \). The remaining rows are evaluated in the same way.

Graph of cos θ against φ = MO/NO

Slope of the graph

Taking two well-separated points on the line of best fit, \( (0.20,\ 0.30) \) and \( (0.52,\ 0.80) \):

\[ s = \frac{\Delta\cos\theta}{\Delta\varphi} = \frac{0.80 - 0.30}{0.52 - 0.20} = \frac{0.50}{0.32} = 1.56 \]

Since \( \cos\theta = n\varphi \), the slope of the line equals the refractive index of the glass, so \( n = 1.56 \).

Two precautions

• The pins were fixed vertically (upright) and their images viewed at eye level to avoid parallax error.
• The pins were kept reasonably far apart (about 4 cm) and a sharp pencil was used so that the traced lines were thin and neat.

(b)(i) Snell's law of refraction

The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media; and the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

\[ \frac{\sin i}{\sin r} = n \]

where \( i \) is the angle of incidence and \( r \) is the angle of refraction.

(b)(ii) Critical angle

For the glass of refractive index \( n = 1.5 \), at the critical angle \( C \) the ray is refracted along the surface \( (r = 90^\circ) \), so:

\[ \sin C = \frac{1}{n} = \frac{1}{1.5} = 0.6667 \]\[ C = \sin^{-1}(0.6667) = 41.8^\circ \]