CBSE Class XII • Unit 06 (Optics)
Chapter 09: Ray Optics and Optical Instruments
Master Revision Dossier • Spherical Mirrors, Curved Refraction, Lens Maker's Formula, Prism Deviation & Optical Instruments
Open Access Initiative
Unit VI Bracket: 14 Marks Core
1. Definitive Formula & Optical Relations Matrix
| Concept | Mathematical Formulation | Sign Conventions | Boundary Conditions & Notes |
|---|---|---|---|
| Mirror Formula | $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = \frac{2}{R}, \quad m = -\frac{v}{u} = \frac{f}{f-u}$$ | Concave: $f < 0$ Convex: $f > 0$ |
Focal length $f = R/2$ is independent of the surrounding medium. Longitudinal magnification $m_L = -m^2$. |
| Apparent Depth & Shift | $$h' = \frac{h}{\mu}, \quad \text{Normal Shift } \Delta t = t\left(1 - \frac{1}{\mu}\right)$$ | Denser into rarer viewing | Shift $\Delta t$ is independent of the distance of the object from the slab. Circle of vision area $A = \frac{\pi d^2}{\mu^2 - 1}$. |
| Critical Angle & TIR | $$\sin i_c = \frac{\mu_R}{\mu_D} = \frac{1}{\mu} \quad (i > i_c \text{ for TIR})$$ | Must go from Denser to Rarer | Optical fibre numerical aperture: $\text{NA} = \sin\theta_a = \sqrt{\mu_1^2 - \mu_2^2}$. 100% reflection without loss. |
| Curved Surface Refraction | $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ | $\mu_1$: Object medium $\mu_2$: Image medium |
Valid for concave or convex surfaces. Light moves from medium 1 towards medium 2 across curvature $R$. |
| Lens Maker's Formula | $$\frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ | Equiconvex: $R_1=+R, R_2=-R$ Equiconcave: $R_1=-R, R_2=+R$ |
For glass lens in water: $f_w = 4 f_a$. If immersed in liquid with $\mu_l > \mu_g$, lens nature reverses (convex becomes diverging). |
| Thin Lens Equation | $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \quad m = +\frac{v}{u} = \frac{f}{f+u}$$ | Convex: $f > 0$ Concave: $f < 0$ |
Optical power $P = \frac{1}{f(\text{m})}$ in Dioptres (D). Combination in contact: $P_{\text{eq}} = P_1 + P_2 \implies \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. |
| Prism Formula | $$\delta = i + e - A, \quad \mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ | At $D_m$: $i = e, r_1 = r_2 = A/2$ | Thin prism: $\delta = (\mu - 1)A$. Dispersive power $\omega = \frac{\mu_V - \mu_R}{\mu_Y - 1}$ is independent of prism angle $A$. |
| Compound Microscope | $$m_{\text{near}} = -\frac{L}{f_o}\left(1 + \frac{D}{f_e}\right), \quad m_{\infty} = -\frac{L}{f_o}\left(\frac{D}{f_e}\right)$$ | $D = 25\text{ cm}$ $f_o, f_e \text{ both small}$ |
Total magnification $m = m_o \times m_e$. Normal adjustment tube length $L = v_o + f_e$. Near point tube length $L = v_o + u_e$. |
| Astronomical Telescope | $$m_{\infty} = -\frac{f_o}{f_e} \quad (L = f_o + f_e)$$ $$m_{\text{near}} = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) \quad (L = f_o + u_e)$$ | $f_o \text{ large, } f_e \text{ small}$ Aperture $D$ large |
Resolving power $\text{RP} = \frac{D}{1.22\lambda} \propto D$. Normal adjustment gives relaxed eye viewing with final image at infinity. |
2. Core Derivations Blueprint (Step-by-Step Board Proofs)
Blueprint 9.1: The Mirror Equation ($\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$) for a Concave Mirror
1. Consider an object $AB$ placed beyond $C$ on the principal axis of a concave mirror of small aperture, forming real inverted image $A'B'$ between $C$ and $F$.
2. From similar right-angled triangles $\triangle A'B'P$ and $\triangle ABP$: $\frac{A'B'}{AB} = \frac{B'P}{BP}$.
3. From similar triangles $\triangle A'B'F$ and $\triangle MPF$ (where $MP \perp \text{axis}$ is incident aperture height, $MP = AB$): $\frac{A'B'}{MP} = \frac{A'B'}{AB} = \frac{B'F}{FP} = \frac{B'P - FP}{FP}$.
4. Equating both expressions: $\frac{B'P}{BP} = \frac{B'P - FP}{FP}$.
5. Applying Cartesian Sign Convention: $BP = -u$, $B'P = -v$, $FP = -f$:
6. Dividing throughout by $v$:
Blueprint 9.2: Refraction at a Single Spherical Curved Surface ($\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$)
1. Point object $O$ in medium $\mu_1$ forms image $I$ in medium $\mu_2$ across curved interface of radius $R$ and center $C$.
2. Let ray $ON$ make angle $\alpha$ with axis, refracted ray $NI$ make angle $\beta$, and normal $NC$ make angle $\gamma$.
3. Exterior angles: In $\triangle ONC$, $i = \alpha + \gamma$. In $\triangle NIC$, $\gamma = r + \beta \implies r = \gamma - \beta$.
4. By Snell's Law for paraxial rays ($\mu_1 i = \mu_2 r$): $\mu_1(\alpha + \gamma) = \mu_2(\gamma - \beta) \implies \mu_1\alpha + \mu_2\beta = (\mu_2 - \mu_1)\gamma$.
5. Substituting paraxial angles $\alpha \approx \frac{NM}{-u}$, $\beta \approx \frac{NM}{v}$, $\gamma \approx \frac{NM}{R}$:
Blueprint 9.3: Deduction of the Lens Maker's Formula
1. For refraction at first surface (radius $R_1$, medium $\mu_1$ to $\mu_2$, intermediate image at $v_1$):
2. For refraction at second surface (radius $R_2$, medium $\mu_2$ to $\mu_1$, virtual object at $v_1$, final image at $v$):
3. Adding Eq. 1 and Eq. 2 ($\frac{\mu_2}{v_1}$ cancels): $\frac{\mu_1}{v} - \frac{\mu_1}{u} = (\mu_2 - \mu_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Dividing by $\mu_1$:
Blueprint 9.4: The Master Prism Formula ($\mu = \frac{\sin\frac{A + D_m}{2}}{\sin\frac{A}{2}}$)
1. Angle of deviation: $\delta = (i - r_1) + (e - r_2) = (i + e) - (r_1 + r_2)$.
2. In quadrilateral $AQOR$: $A + \angle QOR = 180^\circ$. In $\triangle QOR$: $r_1 + r_2 + \angle QOR = 180^\circ \implies r_1 + r_2 = A$.
3. Therefore: $\delta = i + e - A \implies i + e = A + \delta$.
4. At minimum deviation ($D_m$): $i = e$ and $r_1 = r_2 = r \implies 2r = A \implies r = A/2$. Also, $2i = A + D_m \implies i = \frac{A + D_m}{2}$.
5. Applying Snell's Law $\mu = \frac{\sin i}{\sin r}$:
3. High-Yield "Red Alert" Exam Trap Matrix
Master Trap 1: Physical Modifications across Optical Elements
| Physical Action | Effect on Focal Length ($f$) | Effect on Image Formation | Physics Rationale |
|---|---|---|---|
| Mirror immersed in water | No Change ($f' = f$) | Complete image, same size | $f = R/2$ is pure geometry; reflection laws are medium-independent. |
| Glass lens immersed in water | Increases 4 times ($f_w = 4 f_a$) | Power drops to 25% | Relative index drops: $(\mu_g/\mu_w - 1) = (9/8 - 1) = 1/8$ vs $(1.5 - 1) = 1/2$. |
| Glass lens in liquid ($\mu_l > \mu_g$) | Sign Reverses ($f < 0$) | Convex acts as Diverging | $(\mu_g/\mu_l - 1)$ becomes negative; convergence/divergence inverts. |
| Lens cut transversely ($\perp$ axis) | Doubles ($f' = 2f$) | Power halves ($P' = P/2$) | One curved face becomes flat ($R_2 \to \infty$); plano-convex lens forms. |
| Lens cut longitudinally ($\parallel$ axis) | No Change ($f' = f$) | Intensity drops to 50% | Curvatures $R_1, R_2$ unchanged, but aperture collection area halves. |
| Half aperture covered black | No Change | Full Image Still Formed! | Every uncovered point forms full image; only total brightness reduces. |
⚠️ Trap 2: Microscope vs Telescope Lenses Swap
A telescope CANNOT be converted into a microscope by reversing lenses:
• Microscope requires $f_o < f_e$, both very small ($\sim \text{mm}$ to few $\text{cm}$).
• Telescope requires $f_o \gg f_e$ ($f_o \sim 100\text{ cm}$, large aperture for light gathering).
⚠️ Trap 3: Telescope Magnification vs Resolving Power
Magnification $m = f_o/f_e$ only increases angular size.
Ability to distinguish close stars is Resolving Power ($\text{RP} = \frac{D}{1.22\lambda}$), which depends strictly on objective aperture diameter $D$, not on focal lengths!
4. Handcrafted Exemplars (Analytical Numerical Solutions)
A convex lens of glass ($\mu_g = 1.5$) has focal length $15\text{ cm}$ in air. It is immersed in a liquid of refractive index $\mu_l = 1.65$. Calculate its focal length in the liquid and state its optical nature.
Solution: In air: $\frac{1}{f_a} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies \frac{1}{15} = (1.5 - 1)\Delta R \implies \Delta R = \frac{1}{7.5}\text{ cm}^{-1}$.
In liquid ($\mu_l = 1.65$):
An astronomical telescope has an objective of focal length $100\text{ cm}$ and an eyepiece of focal length $5\text{ cm}$. Find: (i) Magnifying power in normal adjustment, (ii) Separation between lenses, and (iii) Magnifying power when the final image is formed at the near point ($D = 25\text{ cm}$).
Solution: $f_o = 100\text{ cm}, f_e = 5\text{ cm}, D = 25\text{ cm}$.
5. Printable Pocket Flashcard Grid
Mirror Law
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Lens Law
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Prism Formula
$\mu = \frac{\sin\frac{A+D_m}{2}}{\sin\frac{A}{2}}$
Telescope Power
$m = -\frac{f_o}{f_e}$ ($L=f_o+f_e$)