Question

Derive an expression for refractive index of an equilateral prism in terms of the angle of minimum deviation and the angle of the prism. Plot a graph to show the variation of the angle of deviation as a function of angle of incidence.

Hint:

Use r1 + r2 = A and D = i + e - A. At minimum deviation i = e and r = A/2, so n = sin((A + Dm)/2) / sin(A/2). The D-i graph is a U-shaped curve.

Answer 1

Step 1: Understanding the Concept:
A ray of light passing through a prism is bent twice, once at each face. The angle between the incident ray and the emergent ray is the angle of deviation \(D\). It depends on the angle of incidence \(i\).

Step 2: Set up the diagram (described in words):
Draw a triangular prism ABC with the angle of the prism \(A\) at the top vertex. A ray PQ meets face AB at Q with angle of incidence \(i\) and refracts along QR at angle \(r_1\). It meets face AC at R at angle \(r_2\) inside the prism, and emerges along RS at angle \(e\). Extend the incident and emergent rays to meet. The angle between them is \(D\).

Step 3: Relations from geometry:
In the quadrilateral AQNR, the angles at Q and R are right angles (normals to the faces). So:
\[ r_1 + r_2 = A \]
The total deviation is the sum of the deviations at the two faces, \((i - r_1)\) and \((e - r_2)\):
\[ D = i + e - (r_1 + r_2) = i + e - A \]

Step 4: Condition for minimum deviation:
The deviation is least when the ray passes symmetrically through the prism. Then \(i = e\) and \(r_1 = r_2 = r\). From the relations above:
\[ r = \frac{A}{2}, \qquad D_m = 2i - A \ \Rightarrow\ i = \frac{A + D_m}{2} \]

Step 5: Apply Snell's law:
\[ n = \frac{\sin i}{\sin r} = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
For an equilateral prism \(A = 60^{\circ}\) and \(\sin 30^{\circ} = \frac{1}{2}\). So:
\[ n = 2\sin\left(\frac{60^{\circ} + D_m}{2}\right) \]

Step 6: The graph of D against i (described in words):
Draw \(i\) on the horizontal axis and \(D\) on the vertical axis. The curve starts at a high value of \(D\) at a small angle of incidence, falls to a minimum \(D_m\) at \(i = e\), and rises again for larger \(i\). It is a smooth U-shaped curve. For each \(D\) above \(D_m\) there are two values of \(i\) (the second one is the value of \(e\) of the first, by the reversibility of light).

Final Answer:
The refractive index of the prism material is: \[ \boxed{n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2\sin\left(\frac{60^{\circ} + D_m}{2}\right)} \]