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AXIOM PHYSICS Class 12 • Module 9.4

Unit 06: Optics • Chapter 09 (Ray Optics)

Prism Refraction, Minimum Deviation & Dispersion

Geometry of deviation ($\delta = i + e - A$), symmetry condition at $D_m$, prism refractive index formula, thin prism deviation, and angular dispersion.

⚡ 01. The Spark (Analogy)

Socho ek rectangular glass slab ko: jab light usme enter karti hai, toh pehle surface par ek taraf mudti hai aur doosre parallel surface par theek utna hi ulti taraf mudkar wapas parallel nikal jaati hai (zero net deviation, only lateral shift). Lekin ek Prism mein refracting surfaces parallel nahi hoti, balki ek angle ($A$) par inclined hoti hain! Yahan dono surfaces par light ka bend ek hi direction mein (towards the thicker base) add up hota hai, jisse incoming ray ek permanent Angle of Deviation ($\delta$) par turn ho jaati hai. Aur kyunki glass ke andar alag-alag wavelengths ki speed alag hoti hai (Cauchy's relation: $\mu_{\text{violet}} > \mu_{\text{red}}$), violet light sabse zyada bend hoti hai aur red sabse kam—jisse ordinary white light apne andar chhipe saat rangon (VIBGYOR) ke spectrum mein bikharka nikal aati hai!

📐 02. The Core Law & Formulae

1. General Prism Deviation & Prism Formula

Relations for prism angle $A$, incidence $i$, and emergence $e$:

$$r_1 + r_2 = A, \qquad \delta = i + e - A$$ $$\text{At Minimum Deviation } (D_m): \quad i = e, \quad r_1 = r_2 = \frac{A}{2}$$ $$\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Symmetry Rule: At minimum deviation, the refracted ray inside the prism travels strictly parallel to the base of an equilateral or isosceles prism.

2. Thin Prism & Dispersive Power

For small prism angle ($A \le 10^\circ$) and wavelength dispersion:

$$\delta = (\mu - 1)A$$ $$\text{Angular Dispersion: } \theta = \delta_V - \delta_R = (\mu_V - \mu_R)A$$ $$\text{Dispersive Power: } \omega = \frac{\theta}{\delta_Y} = \frac{\mu_V - \mu_R}{\mu_Y - 1} \quad \left(\mu_Y = \frac{\mu_V + \mu_R}{2}\right)$$

Dispersive power $\omega$ depends strictly on the material of the prism, independent of the angle of prism $A$.

03. Interactive Visualizer: Prism Ray Tracer & $i - \delta$ Dynamic Response
Angle of Incidence ($i$) 48.5°
Prism Angle ($A$) 60.0° (Equilateral)
Refractive Index ($\mu$) 1.52 (Crown Glass)
Deviation Angle ($\delta$):
37.20°
Minimum Deviation ($D_m$):
37.04°
Angle of Emergence ($e$):
48.70°
Symmetry Status:
Near Minimum (i ≈ e)
📝 04. Comprehensive Stepwise Derivations Blueprint
Primary Board Derivation

Deduction of the Angle of Deviation ($\delta = i + e - A$) and the Master Prism Formula

A B C P (Incident) Q N₁ R N₂ O S (Emergent) T δ i r₁ r₂ e

Fig 9.4A: Ray traversal through a triangular prism showing incident ray $PQ$, internal refracted ray $QR$, emergent ray $RS$, and angle of deviation $\delta$.

1. Consider a ray $PQ$ incident at angle $i$ on refracting face $AB$ of a glass prism with refracting angle $A$. The ray refracts at angle $r_1$ along $QR$, and hits the opposite face $AC$ at angle of incidence $r_2$. It emerges into air at angle of emergence $e$ along $RS$.

2. At face $AB$, the ray deviates clockwise by angle $\delta_1 = i - r_1$. At face $AC$, it further deviates clockwise by angle $\delta_2 = e - r_2$.

3. In $\triangle TQR$, the net angle of deviation $\delta$ is the exterior angle:

$$\delta = \delta_1 + \delta_2 = (i - r_1) + (e - r_2) = (i + e) - (r_1 + r_2) \quad \text{--- [Eq. 1]}$$

4. In cyclic quadrilateral $AQOR$, since normals are perpendicular to the faces ($\angle AQO = \angle ARO = 90^\circ$):

$$\angle A + \angle QOR = 180^\circ \quad \text{--- [Eq. 2]}$$

5. In $\triangle QOR$: $r_1 + r_2 + \angle QOR = 180^\circ$. Comparing with Eq. 2:

$$r_1 + r_2 = A \quad \text{--- [Eq. 3]}$$

6. Substituting Eq. 3 into Eq. 1 yields the universal deviation relation:

$$\delta = i + e - A \implies i + e = A + \delta$$

7. Minimum Deviation Condition ($D_m$): Experimentally, as $i$ increases from small values, $\delta$ first decreases, reaches a unique minimum $D_m$, and then increases again. By the Principle of Reversibility of light, the path must be symmetrical at minimum deviation:

$$i = e \quad \text{and} \quad r_1 = r_2 = r$$

Substituting into $r_1 + r_2 = A$ and $\delta = i + e - A$:

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

8. Applying Snell's Law ($\mu = \frac{\sin i}{\sin r}$) at the first face:

$$\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \quad (\text{Master Prism Formula})$$
Analytical Deduction

Deviation in a Thin Prism ($A \le 10^\circ$), Angular Dispersion & Dispersive Power

1. For a thin prism of small refracting angle $A$, all angles of incidence and refraction are small, allowing the paraxial approximation $\sin\theta \approx \theta$ in Snell's Law:

$$i \approx \mu r_1, \qquad e \approx \mu r_2$$

2. Substituting into the deviation formula $\delta = i + e - A$:

$$\delta = \mu r_1 + \mu r_2 - A = \mu (r_1 + r_2) - A = \mu A - A = (\mu - 1)A$$

3. Angular Dispersion ($\theta$): By Cauchy's formula ($\mu = A' + \frac{B'}{\lambda^2}$), refractive index is higher for shorter wavelengths ($\mu_V > \mu_R$). The angular spread between extreme emergent rays (Violet and Red) is:

$$\theta = \delta_V - \delta_R = (\mu_V - 1)A - (\mu_R - 1)A = (\mu_V - \mu_R)A$$

4. Dispersive Power ($\omega$): The ratio of angular dispersion to the mean deviation (Yellow light $\delta_Y$):

$$\omega = \frac{\theta}{\delta_Y} = \frac{(\mu_V - \mu_R)A}{(\mu_Y - 1)A} = \frac{\mu_V - \mu_R}{\mu_Y - 1}$$

Vital Characteristic: Notice that the prism angle $A$ cancels out completely! Dispersive power is an intrinsic material property of the optical glass, independent of the size or shape of the prism.

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: The Asymmetric Nature of the $i - \delta$ Curve

*Board Question:* Is the $i-\delta$ (angle of incidence vs deviation) curve parabolic and symmetric?
• NO! The curve is ASYMMETRICAL.
• It drops sharply on the left ($i < i_{\min}$) and rises more gradually on the right ($i > i_{\min}$).
• For any deviation value $\delta > D_m$, there are strictly TWO angles of incidence ($i$ and $e$) giving the exact same deviation, but at minimum deviation $D_m$, there is only ONE unique angle ($i = e$).

Trap 2: Limiting Angle of a Prism for No Emergence

*Exam Trap:* What is the maximum angle of a prism ($A$) such that no ray can ever emerge from the second face, regardless of the angle of incidence?
• Even for grazing incidence ($i = 90^\circ$), $r_1 = i_c$.
• For Total Internal Reflection at face 2, we must have $r_2 > i_c \implies A = r_1 + r_2 > 2i_c$.
• If $A > 2i_c$, all incident rays undergo total internal reflection at the second face and NO light ever emerges!

Trap 3: Prism Immersed in Water

When a glass prism of refractive index $\mu_g = 1.5$ is immersed in water ($\mu_w = 1.33$), its relative refractive index drops to $\mu_{\text{rel}} = \frac{1.5}{1.33} \approx 1.125$.
From $\delta = (\mu_{\text{rel}} - 1)A$: The Angle of Deviation decreases substantially (to about 1/4th of its value in air), and angular dispersion is greatly compressed!

Trap 4: Dispersion Without Deviation (Achromatic Pair)

To produce dispersion without net angular deviation, two thin prisms made of crown and flint glass are combined with their refracting angles in opposite directions: $$\delta_{\text{net}} = (\mu_1 - 1)A_1 - (\mu_2 - 1)A_2 = 0 \implies \mathbf{\frac{A_2}{A_1} = \frac{\mu_1 - 1}{\mu_2 - 1}}$$

💡 06. Handcrafted Exemplars
Board Classic Level

Problem: A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle of emergence and each of these angles is equal to $3/4$ of the angle of prism. Calculate (i) the angle of deviation, and (ii) the refractive index of the glass material.

Solution:

1. Equilateral prism: $A = 60^\circ$. Given $i = e = \frac{3}{4}A = \frac{3}{4}(60^\circ) = 45^\circ$.

2. Deviation relation:

$$\delta = i + e - A = 45^\circ + 45^\circ - 60^\circ = 90^\circ - 60^\circ = 30^\circ$$

Since $i = e$, this is the condition of minimum deviation ($D_m = 30^\circ$).

3. Refractive index using master prism formula:

$$\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin 45^\circ}{\sin 30^\circ}$$
$$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$$
JEE/NEET Competitive Trap

Problem: The refractive indices of crown glass for violet and red lights are $1.523$ and $1.514$ respectively. For a thin prism of angle $6^\circ$ made of this glass, calculate (i) the angular dispersion between violet and red, and (ii) the dispersive power of the crown glass.

Solution:

1. Given: $\mu_V = 1.523, \mu_R = 1.514, A = 6^\circ$.

Mean refractive index (Yellow light):

$$\mu_Y = \frac{\mu_V + \mu_R}{2} = \frac{1.523 + 1.514}{2} = \frac{3.037}{2} = 1.5185$$

2. Angular dispersion ($\theta$):

$$\theta = (\mu_V - \mu_R)A = (1.523 - 1.514) \times 6^\circ = 0.009 \times 6^\circ = 0.054^\circ = 3.24\text{ arcminutes}$$

3. Dispersive power ($\omega$):

$$\omega = \frac{\mu_V - \mu_R}{\mu_Y - 1} = \frac{1.523 - 1.514}{1.5185 - 1} = \frac{0.009}{0.5185} \approx 0.01735$$
🗂️ 07. Quick Revision Flashcard Core Formulae

Deviation Angle

$\delta = i + e - A$

Prism Formula

$\mu = \frac{\sin\frac{A+D_m}{2}}{\sin\frac{A}{2}}$

Thin Prism ($A \le 10^\circ$)

$\delta = (\mu - 1)A$

Dispersive Power

$\omega = \frac{\mu_V - \mu_R}{\mu_Y - 1}$