You have been provided with a rectangular glass prism, optical pins, and other necessary apparatus. Using the above diagram as a guide, carry out the follow...

Principle: The construction measures quantities proportional to \(\sin i\) and \(\sin r\), where \(i\) is the angle of incidence and \(r\) is the angle of refraction in the glass.

Using the same circle radius \(R\) for every trial:

\[ d=EF=R\sin i \] \[ l=GH=R\sin r \]

Therefore, by Snell’s law,

\[ \frac{d}{l}=\frac{R\sin i}{R\sin r}=\frac{\sin i}{\sin r}=n \]

Thus, a graph of \(d\) on the vertical axis against \(l\) on the horizontal axis should be a straight line passing approximately through the origin. Its gradient is the refractive index \(n\) of the glass relative to air.

Observation table

S/N	\(i\) / °	\(d=EF\) / cm	\(l=GH\) / cm
1	25	0.85	0.55
2	35	1.15	0.75
3	45	1.50	0.95
4	55	1.65	1.10
5	65	1.85	1.20

Graph of \(d\) against \(l\)

The plotted points are close to a straight line. Using a best-fit line through the origin gives a gradient of approximately \(1.54\), so the refractive index of the glass is approximately:

\[ n \approx 1.5 \text{ to } 1.6 \]

The value \(1.56\) is a reasonable graph-reading estimate. However, the stated numerical table does not itself give exactly \(\frac{1.4}{0.9}\); the gradient must be taken from two widely separated points on the drawn best-fit line, not necessarily directly from two raw data points.

Precautions

• Keep the radius of the circle centred at \(O\) the same for all readings, so that \(d\) and \(l\) remain proportional to \(\sin i\) and \(\sin r\).
• Fix all optical pins vertically.
• Keep the pins reasonably far apart and align \(P_3\) and \(P_4\) with the images of \(P_1\) and \(P_2\) without parallax.
• Use a sharp pencil and make thin, accurate ray lines and normal lines.

Snell’s law: For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant:

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

Critical angle for a water–air boundary: Given \(n=\frac{4}{3}\),

\[ \sin C=\frac{1}{n}=\frac{1}{4/3}=\frac{3}{4}=0.75 \] \[ C=\sin^{-1}(0.75)=48.6^\circ \]

Therefore, the critical angle is \(48.6^\circ\), approximately \(49^\circ\).

Examination reminder: The vertical quantity is \(d=R\sin i\) and the horizontal quantity is \(l=R\sin r\). Therefore the gradient is \(\frac{d}{l}=\frac{\sin i}{\sin r}\), which is the refractive index.