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AXIOM PHYSICS Class 12 • Module 9.3

Unit 06: Optics • Chapter 09 (Ray Optics)

Refraction at Spherical Surfaces & Lens Maker's Formula

Curved interface Snell's law, two-surface refraction coupling, Lens Maker's equation, thin lens formula, optical power, and liquid immersion physics.

⚡ 01. The Spark (Analogy)

Socho ek marathon race ko jisme runners pehle ek curved mitti ke maidan (first refracting surface) par daudte hain aur fir aage chalkar jungle (second refracting surface) mein enter karte hain. Pehla curve unki direction ko halka sa mod deta hai, aur wahi nayi direction doosre curve ke liye starting point ban jaati hai! Ek lens asal mein koi single device nahi hai, balki do back-to-back curved refracting surfaces ka sequential relay race hai. Pehli surface jo intermediate image banati hai, wahi doosri surface ke liye virtual object ka kaam karti hai! Lens Maker's Formula kisi optical engineer ka master recipe card hai—jo batata hai ki glass ke raw piece ko kitne curvature ($R_1, R_2$) par grind kiya jaye taaki exact desired focal length $f$ mil sake!

📐 02. The Core Law & Formulae

1. Single Spherical Surface Refraction

Light going from medium $\mu_1$ into curved medium $\mu_2$ of radius $R$:

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

Fundamental invariant: $\frac{\text{Refractive index of image medium}}{v} - \frac{\text{Refractive index of object medium}}{u} = \frac{\Delta\mu}{R}$.

2. Lens Maker's Formula & Thin Lens Law

Coupling both surfaces for a thin lens of index $\mu_2$ in medium $\mu_1$:

$$\frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \quad m = \frac{h_i}{h_o} = +\frac{v}{u}$$

Power $P = \frac{1}{f(\text{in metres})}$ (Dioptres, D). For thin lenses in contact: $P_{\text{eq}} = P_1 + P_2 \implies \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.

03. Interactive Visualizer: Thin Lens Ray Tracer Engine
Object Position ($u$) -26.0 cm
Focal Length Magnitude ($\vert{}f\vert{}$) 12.0 cm
Object Height ($h_o$) 5.0 cm
Image Distance ($v$):
+22.29 cm
Magnification ($m = +v/u$):
-0.86 (Diminished)
Optical Power ($P = 1/f$):
+8.33 Dioptres
Image Nature:
Real & Inverted
📝 04. Comprehensive Stepwise Derivations Blueprint
Primary Board Proof

Refraction at a Single Spherical Curved Surface ($\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$)

Axis Rarer Medium (μ₁) Denser Medium (μ₂ > μ₁) P C O N Normal (NC) I α γ β i r

Fig 9.3A: Refraction at a convex spherical interface of radius $R$ separating medium $\mu_1$ and $\mu_2$.

1. Consider a point object $O$ placed on the principal axis in medium $\mu_1$. Ray $ON$ strikes the interface at $N$, bending towards the normal $NC$ to converge at image point $I$ in medium $\mu_2$.

2. Let $\angle NOC = \alpha$, $\angle NIC = \beta$, and $\angle NCP = \gamma$. By exterior angle theorem of triangles:

$$\text{In } \triangle ONC: \quad i = \alpha + \gamma$$ $$\text{In } \triangle NIC: \quad \gamma = r + \beta \implies r = \gamma - \beta$$

3. Drop perpendicular $NM$ on the axis. For paraxial rays (small angles $\alpha, \beta, \gamma$), $M \approx P$, and tangents equal angles:

$$\alpha \approx \tan\alpha = \frac{NM}{OM} \approx \frac{NM}{-u}, \quad \beta \approx \tan\beta = \frac{NM}{MI} \approx \frac{NM}{v}, \quad \gamma \approx \tan\gamma = \frac{NM}{MC} \approx \frac{NM}{R}$$

4. Applying Snell's Law for small angles ($\mu_1 \sin i \approx \mu_2 \sin r \implies \mu_1 i = \mu_2 r$):

$$\mu_1 (\alpha + \gamma) = \mu_2 (\gamma - \beta) \implies \mu_1 \alpha + \mu_2 \beta = (\mu_2 - \mu_1)\gamma$$

5. Substituting values of $\alpha, \beta, \gamma$:

$$\mu_1 \left(\frac{NM}{-u}\right) + \mu_2 \left(\frac{NM}{v}\right) = (\mu_2 - \mu_1)\left(\frac{NM}{R}\right)$$
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
Primary Board Derivation

Deduction of the Lens Maker's Formula for a Double Convex Lens

1. Consider a thin double convex lens of refractive index $\mu_2$ placed in a surrounding medium of refractive index $\mu_1$. The radii of curvature of the two surfaces are $R_1$ and $R_2$.

2. Refraction at First Surface ($R_1$): Object $O$ is at distance $u$. Light travels from medium $\mu_1$ to $\mu_2$. It forms an intermediate image $I_1$ at distance $v_1$:

$$\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1} \quad \text{--- [Eq. 1]}$$

3. Refraction at Second Surface ($R_2$): The intermediate image $I_1$ acts as a virtual object for the second surface (object distance $= v_1$). Light travels from lens medium $\mu_2$ back into surrounding medium $\mu_1$. The final image is formed at distance $v$:

$$\frac{\mu_1}{v} - \frac{\mu_2}{v_1} = \frac{\mu_1 - \mu_2}{R_2} = -\frac{\mu_2 - \mu_1}{R_2} \quad \text{--- [Eq. 2]}$$

4. Adding Eq. 1 and Eq. 2 (the intermediate term $\frac{\mu_2}{v_1}$ cancels out identically):

$$\frac{\mu_1}{v} - \frac{\mu_1}{u} = (\mu_2 - \mu_1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

5. Dividing throughout by $\mu_1$:

$$\frac{1}{v} - \frac{1}{u} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \text{--- [Eq. 3]}$$

6. If the object is at infinity ($u = \infty$), by definition the image is formed at the principal focus ($v = f$):

$$\frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad (\text{Lens Maker's Formula})$$

7. Comparing Eq. 3 and the Lens Maker's Formula yields the classical Thin Lens Equation:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Immersion in Liquid ($\mu_{\text{liquid}} > \mu_{\text{glass}}$)

*Board Question:* A glass convex lens ($\mu_g = 1.5$) is placed in carbon disulphide ($\mu_l = 1.65$). How does it behave?
• Relative index: $\frac{\mu_g}{\mu_l} - 1 = \frac{1.5}{1.65} - 1 < 0$ (Negative!).
• The sign of focal length reverses! The convex lens turns into a DIVERGING (concave) lens!
• If $\mu_l = \mu_g$: $1/f = 0 \implies f \to \infty$. The lens becomes completely invisible and acts as a plain glass plate.

Trap 2: Cutting a Lens into Halves (Transverse vs Longitudinal)

• Transverse cut (perpendicular to principal axis): Creates two plano-convex lenses. $R_2 \to \infty \implies \frac{1}{f'} = (\mu-1)\frac{1}{R_1} \implies \mathbf{f' = 2f, P' = P/2}$.
• Longitudinal cut (along principal axis): Curvatures remain identical ($R_1, R_2$). Focal length is UNCHANGED ($f' = f$)! However, aperture area is halved, so intensity drops to 50%.

Trap 3: Sign Convention of $R_1$ and $R_2$

Never make sign errors in Lens Maker's formula:
• Equiconvex Lens: $R_1 = +R$ (center to right), $R_2 = -R$ (center to left). $$\frac{1}{f} = (\mu - 1)\left[\frac{1}{+R} - \frac{1}{-R}\right] = \mathbf{\frac{2(\mu - 1)}{R}} \implies f = \frac{R}{2(\mu - 1)}$$ • Equiconcave Lens: $R_1 = -R$, $R_2 = +R \implies f = -\frac{R}{2(\mu - 1)}$.

Trap 4: Water Bubble Inside Glass / Liquid

An air bubble inside water or glass has a convex geometric shape.
However, light enters from denser medium ($\mu_w = 1.33$) into rarer air medium ($\mu_a = 1.0$).
Hence, an air bubble inside water behaves as a DIVERGING LENS (Concave behavior)!

💡 06. Handcrafted Exemplars
Board Classic Level

Problem: A biconvex lens has a focal length of $20\text{ cm}$ in air. Its glass has refractive index $\mu_g = 1.5$. Calculate its focal length when it is completely immersed in water of refractive index $\mu_w = 4/3$.

Solution:

1. In air ($\mu_{\text{air}} = 1$):

$$\frac{1}{f_a} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies \frac{1}{20} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{10}\text{ cm}^{-1}$$

2. When immersed in water ($\mu_w = 4/3$):

$$\frac{1}{f_w} = \left(\frac{\mu_g}{\mu_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1.5}{4/3} - 1\right)\left(\frac{1}{10}\right) = \left(\frac{9}{8} - 1\right)\left(\frac{1}{10}\right) = \left(\frac{1}{8}\right)\left(\frac{1}{10}\right) = \frac{1}{80}$$
$$f_w = 80\text{ cm} = 4 \times f_a$$

Vital General Result: For glass ($\mu = 1.5$) immersed in water ($\mu = 1.33$), focal length always increases by a factor of exactly 4 times ($f_w = 4 f_a$).

JEE/NEET Competitive Trap

Problem: A converging lens of focal length $30\text{ cm}$ is placed in contact coaxially with a diverging lens of focal length $20\text{ cm}$. Find the power and focal length of the combination. Does the system act as a converging or diverging system?

Solution:

1. Converging lens: $f_1 = +30\text{ cm} = +0.30\text{ m}$. Diverging lens: $f_2 = -20\text{ cm} = -0.20\text{ m}$.

$$P_1 = \frac{1}{+0.30} = +3.33\text{ D}, \qquad P_2 = \frac{1}{-0.20} = -5.00\text{ D}$$

2. Equivalent power of the combination:

$$P = P_1 + P_2 = +3.33\text{ D} + (-5.00\text{ D}) = -1.67\text{ Dioptres}$$

3. Equivalent focal length ($F$):

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{+30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60} \implies F = -60\text{ cm}$$

Since $F < 0$ (and $P < 0$), the combination behaves as a Diverging (Concave) System!

🗂️ 07. Quick Revision Flashcard Core Formulae

Curved Interface

$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$

Lens Maker's Law

$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Thin Lens Law

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ ($m = \frac{v}{u}$)

Liquid Immersion

$f_w = 4 f_a$ (Glass in Water)