(a) Refractive Index of a Triangular Prism

Step 1 (Governing Laws & Geometry): [1 Mark]

For a prism with angle $A$, angle of incidence $i$, angle of emergence $e$, and angle of deviation $\delta$, the following relations hold:

• $A = r_1 + r_2$
• $\delta = (i + r_1) + (e - r_2) = i + e - A$

At the condition of minimum deviation ($\delta = \delta_m$):

• $i = e$
• $r_1 = r_2 = r$

Step 2 (Mathematical Derivation): [3 Marks]

Substituting the minimum deviation conditions into the geometric relations:

$$A = r + r \implies r = \frac{A}{2}$$ $$\delta_m = i + i - A \implies i = \frac{A + \delta_m}{2}$$

Applying Snell's Law at the first interface:

$$n = \frac{\sin i}{\sin r}$$

Substituting the values of $i$ and $r$:

$$n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Step 3 (Condition for Refraction & Final Answer): [1 Mark]

For refraction to take place at the second face, the angle of incidence at the second face ($r_2$) must be less than the critical angle ($C$). Since $r_2 = A/2$, the condition is:

$$\frac{A}{2} < C \implies \sin\left(\frac{A}{2}\right) < \frac{1}{n} \implies n < \frac{1}{\sin(A/2)}$$

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(b) Huygens Principle and Snell's Law

Step 1 (Huygens Principle & Diagram Setup): [1 Mark]

Huygens Principle: Each point on a wavefront acts as a source of secondary wavelets, and the new wavefront is the forward envelope of these wavelets at a later time.

[Diagram Description: A plane wavefront $AB$ incident on interface $XY$. $A$ reaches the interface first, creating a wavelet of radius $v_2t$ in the denser medium while $B$ travels distance $v_1t$ to reach $B'$. The tangent from $B'$ to the wavelet at $A$ forms the refracted wavefront $A'B'$.]

Step 2 (Mathematical Derivation): [3 Marks]

In $\triangle AB'A'$ and $\triangle AB'B$:

• $i = \angle BAB'$ (Angle of incidence)
• $r = \angle ABA'$ (Angle of refraction)

From the geometry:

$$\sin i = \frac{BB'}{AB'} = \frac{v_1 t}{AB'}$$ $$\sin r = \frac{AA'}{AB'} = \frac{v_2 t}{AB'}$$

Dividing the two equations:

$$\frac{\sin i}{\sin r} = \frac{v_1 t}{v_2 t} = \frac{v_1}{v_2}$$

Step 3 (Final Concluded Answer): [1 Mark]

Since the refractive index $n = c