Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

For an equilateral prism, the angle of prism is $A = 60^\circ$. When a ray is incident normally on one face, the angle of incidence at the first surface is $i_1 = 0^\circ$, which implies the angle of refraction $r_1 = 0^\circ$. Consequently, the angle of incidence at the second face is $r_2 = A = 60^\circ$.

The condition for the emergent ray to graze the surface is that the angle of emergence $e = 90^\circ$. We use Snell's Law at the second interface between the prism (refractive index $\mu_p$) and the surrounding medium (refractive index $\mu_m$):

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

• Given: $r_2 = 60^\circ$ and $e = 90^\circ$.
• Substitute these values into Snell's Law:
$$\mu_p \sin(60^\circ) = \mu_m \sin(90^\circ)$$
• Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ and $\sin(90^\circ) = 1$:
$$\mu_p \left( \frac{\sqrt{3}}{2} \right) = \mu_m (1)$$
• Rearranging to solve for the refractive index of the medium ($\mu_m$):
$$\mu_m = \frac{\sqrt{3}}{2} \mu_p$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The refractive index of the medium is:

💡 EL10 Concept Insight: Imagine the light trying to escape the prism; because the prism is tilted at 60 degrees, the light hits the exit wall at a very sharp angle. By changing the surrounding liquid, we make it just "slippery" enough that the light can't break through the surface and instead slides perfectly along the edge.