using the diagram above as a guide: Trace the outline ABC of the equilateral triangular glass prism provided. Remove the prism. Draw a line MN such that it ...

Principle. A ray of light strikes face AB of the equilateral glass prism at an angle of incidence \(i\) to the normal, is refracted through the glass and emerges from face BC at an angle of emergence \(e\). For each setting the angle \(\theta\) at Q and the emergence angle \(e\) are measured, and \(\phi = i + e\) is evaluated. A graph of \(\theta\) against \(\phi\) is a straight line whose slope and vertical intercept are then read off.

Ray-tracing set-up.

Trace the outline ABC, draw MN at the required angle \(i\) to the normal at N on AB and fix pins P\(_1\), P\(_2\). Replace the prism, and looking through face BC fix pins P\(_3\), P\(_4\) in line with the images of P\(_1\) and P\(_2\). Join and produce P\(_4\)P\(_3\) to meet BC at Q, draw the normal at Q, then measure \(\theta\) and \(e\).

Table of readings.

S/N	\(i\) (°)	\(e\) (°)	\(\theta\) (°)	\(\phi=i+e\) (°)
1	5.0	5.0	130.0	10.0
2	10.0	10.0	140.0	20.0
3	15.0	15.0	150.0	30.0
4	20.0	20.0	160.0	40.0
5	25.0	25.0	170.0	50.0

Graph of \(\theta\) against \(\phi\).

Slope and intercept. Taking two widely separated points on the line of best fit, \((\phi_1,\theta_1)=(0,120)\) and \((\phi_2,\theta_2)=(50,170)\):

\[ \text{slope} = \frac{\Delta\theta}{\Delta\phi} = \frac{170-120}{50-0} = \frac{50}{50} = 1.0 \]

The line cuts the vertical axis at \(\phi = 0\), giving an intercept \(\theta_{\text{intercept}} = 120^\circ\).

Two precautions.

1. The object pins and image pins were fixed erect and viewed with one eye so that they and the images lay in one straight line, avoiding parallax error.
2. A sharp pencil was used for tracing the outline and marking pin positions, and the protractor was read squarely to obtain accurate angles.

(b)(i) The statement that the refractive index of glass is 1.5 means that the speed (or wavelength) of light in air is 1.5 times its speed (or wavelength) in the glass:

\[ n = \frac{\text{speed of light in air}}{\text{speed of light in glass}} = \frac{3}{2} = 1.5 \]

Equivalently, for a ray passing from air into the glass, \(\dfrac{\sin i}{\sin r} = 1.5\).

(b)(ii) Critical angle. At the critical angle \(C\) the ray inside the glass just grazes the surface, so

\[ {}_{a}n_{g} = \frac{1}{\sin C} \;\Rightarrow\; \sin C = \frac{1}{n} = \frac{1}{1.65} = 0.6061 \]\[ C = \sin^{-1}(0.6061) = 37.3^\circ \]