Axiom Physics | Class 12 - Refraction, Apparent Depth & TIR

⚡ 01. The Spark (Analogy)

Socho ek four-wheeler jeep ko jo asphalt road se achanak ek kachhi ret (sand) wali zameen par tirchi (oblique) enter karti hai: jo pahiya pehle ret mein jata hai woh slow ho jata hai, jabki road par chal raha doosra pahiya tez rehta hai, jisse jeep apne normal raste se mud (bend) jaati hai! Theek yahi cheez light ke sath hoti hai: light rarer medium (tez speed) se denser medium (kam speed) mein aane par normal ki taraf jhuk jaati hai. Aur socho jab jeep ulta ret se road par jaati hai—woh normal se door mudti hai. Agar aap aane ka angle itna badha dein ki gaadi wapas ret mein hi ghoom kar palat jaye, toh wahi hai Total Internal Reflection (TIR)! Duniya bhar ka high-speed internet isi TIR par tike patle glass ke baalon (optical fibres) ke zariye light speed par travel kar raha hai!

📐 02. The Core Law & Formulae

1. Snell’s Law & Apparent Depth

Refraction across planar boundary separating media 1 and 2:

$$\mu_1 \sin i = \mu_2 \sin r \implies \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} = {}^{1}\mu_2 = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}$$

$$h_{\text{apparent}} = \frac{h_{\text{real}}}{\mu}, \qquad \text{Normal Shift } \Delta t = t \left(1 - \frac{1}{\mu}\right)$$

Frequency ($\nu$) remains strictly invariant across refraction! Speed and wavelength change in direct proportion.

2. Critical Angle & Total Internal Reflection

When light travels from Denser ($\mu_D$) to Rarer ($\mu_R$):

$$\sin i_c = \frac{\mu_R}{\mu_D} \implies \text{For Denser into Air: } \sin i_c = \frac{1}{\mu}$$ $$\text{Condition for TIR: } i > i_c \quad (\text{100\% Light is Reflected; Zero Loss!})$$

Unlike silvered mirrors which absorb 5-10% of light, TIR reflects 100% of incident radiant energy without attenuation.

03. Interactive Visualizer: Denser-to-Rarer Ray Tracing & TIR Simulator

MODE: PARTIAL REFRACTION

Angle of Incidence in Denser Medium ($i$) 35.0°

Refractive Index of Denser Medium ($\mu$) 1.50 (Crown Glass)

Critical Angle ($i_c = \sin^{-1}(1/\mu)$):

41.81°

Refraction Angle in Air ($r$):

59.36°

Reflection Status:

Partial Reflection (4%)

Energy Efficiency:

Refracted Light Dominates

📝 04. Comprehensive Stepwise Derivations Blueprint

Classic Board Derivation

Apparent Depth of an Object Immersed in Denser Medium & Normal Shift ($\Delta t$)

Fig 9.2A: Geometric formation of virtual image $I$ above real object $O$ due to refraction away from normal.

1. Consider an object $O$ situated at real depth $h = OA$ inside a medium of refractive index $\mu$. A ray $OA$ incident normally passes undeviated.

2. A second ray $OB$ incident at angle $i$ refracts into air at angle $r$ ($r > i$). Produced backwards, it meets the normal line at image point $I$. The apparent depth is $h' = IA$.

3. In right-angled $\triangle OAB$: $\sin i = \frac{AB}{OB}$. In right-angled $\triangle IAB$: $\sin r = \frac{AB}{IB}$.

4. Applying Snell's law ($\mu \sin i = 1 \times \sin r$):

$$\mu = \frac{\sin r}{\sin i} = \frac{AB/IB}{AB/OB} = \frac{OB}{IB}$$

5. For near-normal viewing (paraxial rays, small angles $i$ and $r$), point $B$ lies very close to $A$ ($OB \approx OA = h$, and $IB \approx IA = h'$):

$$\mu = \frac{OA}{IA} = \frac{h_{\text{real}}}{h_{\text{apparent}}} \implies h_{\text{apparent}} = \frac{h}{\mu}$$

6. The upward vertical displacement of the object is known as the Normal Shift ($\Delta t$):

$$\Delta t = h - h' = h - \frac{h}{\mu} = h\left(1 - \frac{1}{\mu}\right)$$

Multiple Slabs Generalization: For $n$ layers of thicknesses $t_1, t_2, \dots$ and indices $\mu_1, \mu_2, \dots$: $\Delta t_{\text{total}} = \sum t_i \left(1 - \frac{1}{\mu_i}\right)$.

Boundary Condition Proof

Deduction of Critical Angle ($i_c$) & Acceptance Angle in Optical Fibres

1. When a ray travels from a denser medium ($\mu_1 = \mu$) to a rarer medium ($\mu_2 = 1$), it bends away from the normal ($r > i$).

2. As angle of incidence $i$ increases, angle of refraction $r$ increases correspondingly until at a specific angle $i = i_c$ (Critical Angle), the refracted ray grazes the interface ($r = 90^\circ$). Applying Snell's Law:

$$\mu \sin i_c = 1 \times \sin 90^\circ = 1 \times 1 = 1$$

$$\sin i_c = \frac{1}{\mu} \implies i_c = \sin^{-1}\left(\frac{1}{\mu}\right)$$

3. If $i > i_c$, no refraction is mathematically possible ($\sin r > 1$, which is invalid). The entire light energy is reflected back into the denser medium following laws of reflection. This is Total Internal Reflection (TIR).

4. Optical Fibre Core-Cladding Boundary: Light entering a core of index $\mu_1$ surrounded by cladding of index $\mu_2$ ($\mu_1 > \mu_2$) undergoes repeated TIR if the angle of incidence at core-cladding boundary exceeds $i_c = \sin^{-1}(\mu_2/\mu_1)$. The maximum external entrance angle in air is the Acceptance Angle ($\theta_a$):

$$\sin\theta_a = \sqrt{\mu_1^2 - \mu_2^2} \quad (\text{Numerical Aperture } \text{NA} = \sin\theta_a)$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Diver Underwater Looking at the External World (Circle of Vision)

*Board Question:* A point light source is placed at depth $d$ under water ($\mu$). What is the area of the circular patch on the surface through which light emerges into air?
• Rays with $i \le i_c$ emerge; rays with $i > i_c$ undergo TIR back down.
• Radius of circular disc: $r = d \tan i_c = \frac{d \sin i_c}{\cos i_c} = \frac{d (1/\mu)}{\sqrt{1 - 1/\mu^2}} = \frac{d}{\sqrt{\mu^2 - 1}}$. $$\mathbf{\text{Area } A = \pi r^2 = \frac{\pi d^2}{\mu^2 - 1}}$$

Trap 2: Shift through a Glass Slab vs Position of Object

*Exam Trap:* "If an object is moved further away from a glass slab of thickness $t$, does the normal shift increase?"
*Correct Answer:* NO! IT REMAINS INDEPENDENT! $$\Delta t = t\left(1 - \frac{1}{\mu}\right)$$ Normal shift depends strictly on the slab's thickness $t$ and refractive index $\mu$. It is completely independent of the distance of the object from the slab!

Trap 3: Prism Inversion Without Energy Loss (Porro Prisms)

Why do binoculars use $45^\circ-90^\circ-45^\circ$ right-angled glass prisms instead of plane mirrors?
• For crown glass ($\mu \approx 1.5$), critical angle is $i_c = \sin^{-1}(1/1.5) \approx \mathbf{41.8^\circ}$.
• When light strikes the internal hypotenuse at $45^\circ$, $i = 45^\circ > 41.8^\circ$, causing 100% Total Internal Reflection with zero image degradation, no ghost reflections, and no silver tarnishing over time.

Trap 4: Diamond Sparkle Secret

Diamond has an exceptionally high refractive index ($\mu \approx 2.42$). Its critical angle is extremely small: $$i_c = \sin^{-1}\left(\frac{1}{2.42}\right) \approx \mathbf{24.4^\circ}$$ A skilled gem-cutter cuts diamond facets so that light entering undergoes multiple internal reflections at angles $> 24.4^\circ$ before emerging through top faces, creating dazzling brilliance.

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A small bulb is placed at the bottom of a water tank of depth $80\text{ cm}$. What is the area of the surface of water through which light from the bulb can emerge out? (Refractive index of water $\mu = 4/3$).

Solution: Depth $d = 80\text{ cm} = 0.80\text{ m}$, $\mu = 4/3$.

1. Critical angle condition: $\sin i_c = \frac{1}{\mu} = \frac{3}{4} = 0.75$.

$$\tan i_c = \frac{\sin i_c}{\cos i_c} = \frac{\sin i_c}{\sqrt{1 - \sin^2 i_c}} = \frac{3/4}{\sqrt{1 - 9/16}} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$$

2. Radius of the circular illuminated disc on surface:

$$r = d \tan i_c = (0.80\text{ m}) \times \frac{3}{\sqrt{7}} = \frac{2.4}{\sqrt{7}}\text{ m} \approx 0.907\text{ m}$$

3. Area through which light emerges:

$$\text{Area } A = \pi r^2 = \pi \left(\frac{2.4}{\sqrt{7}}\right)^2 = \frac{5.76\pi}{7} \approx 2.585\text{ m}^2$$

JEE/NEET Competitive Trap

Problem: An optical fibre has a core of refractive index $\mu_1 = 1.68$ and a cladding of index $\mu_2 = 1.44$. What is the critical angle at the core-cladding boundary? Find the range of incident angles for incoming rays in air at the face of the fibre so that total internal reflection occurs inside the core.

Solution:

1. Critical angle at the internal core-cladding boundary:

$$\sin i_c = \frac{\mu_2}{\mu_1} = \frac{1.44}{1.68} = \frac{6}{7} \approx 0.8571 \implies i_c = \sin^{-1}(0.8571) \approx 59.0^\circ$$

2. For TIR inside the core, the angle with normal at core wall must be $i \ge 59.0^\circ$.

Since the entrance face is normal to the core axis, refraction angle at entrance is $r = 90^\circ - i \le 90^\circ - 59^\circ = 31.0^\circ$.

3. Applying Snell's law at entrance face from air ($\mu_{\text{air}} = 1$):

$$1 \times \sin i_{\max} = \mu_1 \sin r_{\max} = 1.68 \times \sin(31.0^\circ) = 1.68 \times \sqrt{1 - \sin^2(59.0^\circ)} = \sqrt{\mu_1^2 - \mu_2^2}$$ $$\sin i_{\max} = \sqrt{(1.68)^2 - (1.44)^2} = \sqrt{2.8224 - 2.0736} = \sqrt{0.7488} \approx 0.8653$$

$$i_{\max} = \sin^{-1}(0.8653) \approx 59.9^\circ \implies \text{Acceptance Range: } 0^\circ \le i \le 59.9^\circ$$

🗂️ 07. Quick Revision Flashcard Core Formulae

Critical Angle

$\sin i_c = \frac{1}{\mu}$

Apparent Depth

$h' = \frac{h}{\mu}$

Normal Shift

$\Delta t = t(1 - \frac{1}{\mu})$

Circle of Vision

$A = \frac{\pi d^2}{\mu^2 - 1}$

Refraction, Apparent Depth & Total Internal Reflection