Question

Derive the relation for the refractive index ($\mu$) of a prism in terms of angle of minimum deviation ($\delta_m$) and angle of prism ($A$).

• A. $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
• B. $\mu = \frac{\cos\left(\frac{A + \delta_m}{2}\right)}{\cos\left(\frac{A}{2}\right)}$
• C. $\mu = \frac{\sin\left(\frac{A - \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
• D. $\mu = \frac{\tan\left(\frac{A + \delta_m}{2}\right)}{\tan\left(\frac{A}{2}\right)}$

Hint:

At the position of minimum deviation ($\delta = \delta_m$), the refracted ray passing through the interior of the prism becomes perfectly parallel to the base of the prism if the prism is isosceles or equilateral.

Answer 1

The Correct Option is A

Concept: When a monochromatic ray of light travels through a triangular glass prism, it undergoes refraction twice—once at the first incident face and once at the second emerging face. The structural orientation of the two faces causes the emergent ray to bend away from its original path. This angular separation between the direction of the incident ray and the emergent ray is known as the angle of deviation ($\delta$). Let us define the parameters involved in a standard triangular prism setup:
• $A$: Angle of the prism (refracting angle).
• $i$: Angle of incidence at the first face.
• $r_1$: Angle of refraction inside the first face.
• $r_2$: Angle of incidence inside the second face.
• $e$: Angle of emergence from the second face.
• $\delta$: Total angle of deviation.

Step 1: Geometrical relations inside the prism.
Consider a ray of light passing through a prism $ABC$. Let the refraction at the first face $AB$ have an angle of incidence $i$ and angle of refraction $r_1$. At the second face $AC$, let the internal angle be $r_2$ and the final angle of emergence be $e$. From the cyclic quadrilateral formed by the normals and the vertices of the prism, the sum of the internal angles satisfies: $$A + \angle N = 180^\circ$$ In the interior triangle formed by the light ray and the normal intersection: $$r_1 + r_2 + \angle N = 180^\circ$$ Equating these two geometric properties yields the fundamental structural relation of a prism: $$A = r_1 + r_2 \quad \cdots (1)$$ Now, considering the total angular deviation $\delta$ suffered by the ray, it is the sum of deviations at both refracting surfaces: $$\delta = (i - r_1) + (e - r_2)$$ Rearranging the terms gives: $$\delta = (i + e) - (r_1 + r_2)$$ Substituting equation (1) into this expression provides the second fundamental prism formula: $$\delta = i + e - A \quad \cdots (2)$$

Step 2: Condition for minimum deviation.
Experimentally and theoretically, as the angle of incidence $i$ is gradually increased, the angle of deviation $\delta$ first decreases, reaches a certain minimum value ($\delta_m$), and then increases. At this unique condition of minimum deviation ($\delta = \delta_m$), the light ray passes completely symmetrically through the prism. Under this perfect symmetric state, we have: $$i = e$$ $$r_1 = r_2 = r$$ Substituting these symmetric conditions back into equations (1) and (2): From equation (1): $$A = r + r = 2r \quad \Rightarrow \quad r = \frac{A}{2} \quad \cdots (3)$$ From equation (2): $$\delta_m = i + i - A = 2i - A$$ $$2i = A + \delta_m \quad \Rightarrow \quad i = \frac{A + \delta_m}{2} \quad \cdots (4)$$

Step 3: Applying Snell's Law to determine the refractive index.
According to Snell's Law at the first refracting boundary interface (moving from air to glass): $$\mu = \frac{\sin i}{\sin r}$$ Substituting the explicit symmetric structural values of $i$ and $r$ derived in equations (3) and (4) directly into Snell's formula: $$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ This is the famous prism formula that determines the exact refractive index of the material, validating choice (A).