Class 12 Physics NCERT REPRINT 2026-27

Chapter 9: Ray Optics & Optical Instruments

Complete concise study notes, spherical mirrors, mirror equation, Snell's law, total internal reflection, optical fibres, refraction at spherical surfaces, lens maker's formula, prisms, microscopes, telescopes, solved examples 9.1–9.8, and exercises 9.1–9.31.

01 / Exam-Focused Notes

Chapter 9: Ray Optics and Optical Instruments

Complete official NCERT textbook coverage with high-precision vector ray diagrams: Mirror Equation ($1/v + 1/u = 1/f$), Focal Length ($f=R/2$), Snell's Law & Refractive Index, Apparent Depth, Total Internal Reflection & Optical Fibres, Lens Maker's Formula, Lens Power ($P=1/f$), Combination of Thin Lenses, Refraction through Prism & Minimum Deviation ($D_m$), Compound Microscope ($m = m_o m_e$), and Astronomical Refracting & Cassegrain Telescopes.

100% SYLLABUS
Chapter Navigation • NCERT Sections 9.1 – 9.7
9.1-9.2Reflection & Mirrors 9.3-9.4Refraction & TIR 9.5Spherical Lenses & Power 9.6Prism Formula 9.7Microscopes & Telescopes
9.1 & 9.2
NCERT Sections

Introduction, Reflection of Light & Spherical Mirrors

Geometric Optics & Cartesian Sign Convention

9.2.1 Ray Approximation, Laws of Reflection & Sign Convention

  • Ray Model: Wavelength of visible light ($\sim 400\text{ nm}$ to $750\text{ nm}$) is negligibly small compared to macroscopic obstacles, allowing light paths to be modeled as straight-line rays and ray bundles as beams.
  • Laws of Reflection:
    1. Angle of incidence equals angle of reflection ($\theta_i = \theta_r$).
    2. Incident ray, reflected ray, and normal to the reflecting surface at the point of incidence lie in the same plane.
  • Cartesian Sign Convention:
    • All distances are measured from the Pole ($P$) of the spherical mirror or Optical Centre ($O$) of the lens along the principal axis.
    • Distances measured in the direction of incident light are positive ($+$); distances measured opposite to the direction of incident light are negative ($-$).
    • Heights measured upward normal to the principal axis ($+y$) are positive; downward heights ($-y$) are negative.
Derivations

9.2.2 Focal Length ($f = R/2$) & The Mirror Equation

  • Derivation of $f = R/2$:
    For paraxial rays incident at angle $\theta$ on spherical mirror of radius $R$: $$\tan\theta \approx \theta = \frac{MD}{CD} = \frac{MD}{R}, \qquad \tan(2\theta) \approx 2\theta = \frac{MD}{FD} = \frac{MD}{f}$$ $$\implies \frac{MD}{f} = 2 \frac{MD}{R} \implies f = \frac{R}{2}$$ (Concave mirror: $f = -R/2$; Convex mirror: $f = +R/2$).
  • Mirror Equation Derivation:
    From similar triangles $A'B'F \sim MPF$ and $A'B'P \sim ABP$: $$\frac{A'B'}{AB} = \frac{B'F}{FP} = \frac{B'P - FP}{FP} = \frac{-v - (-f)}{-f} = \frac{v - f}{f}$$ $$\frac{A'B'}{AB} = \frac{B'P}{BP} = \frac{-v}{-u} = \frac{v}{u}$$ Equating both ratios: $$\frac{v - f}{f} = \frac{v}{u} \implies \frac{v}{f} - 1 = \frac{v}{u} \implies \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = \frac{2}{R}, \qquad m = \frac{h'}{h} = -\frac{v}{u} = \frac{f - v}{f} = \frac{f}{f - u}$$
Fig 9.3: Concave Mirror Ray Diagram (Real, Inverted Image Formation)
P (Pole) C (2f) F (f) A (Object) B A' (Real Image) B' Mirror Formula: 1/v + 1/u = 1/f • m = -v/u
Example 9.1

Covering Half of a Mirror Surface

Suppose that the lower half of the concave mirror's reflecting surface is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?

Analysis: The laws of reflection hold for every individual point on the uncovered top half of the mirror. Light rays from every point of the object still reach the image location.
Effect: The complete image of the entire object is formed without any missing parts, but the total light energy collected is halved, reducing the image intensity (brightness) by half.
Example 9.2

Mobile Phone along Principal Axis & Non-uniform Magnification

A mobile phone lies along the principal axis of a concave mirror. Show by suitable diagram the formation of its image and explain why magnification is non-uniform.

Non-uniform Magnification: Different parts of the phone lie at different object distances $u$. Since longitudinal magnification $m_L \propto m_T^2 = \left(\frac{f}{u - f}\right)^2$, the end closer to the focus $F$ suffers significantly greater magnification than the end closer to the centre of curvature $C$.
Distortion: The image is distorted into a tapered shape and depends on phone position relative to $F$ and $C$.
Example 9.3

Image Position & Nature for Concave Mirror

An object is placed at (i) $10\text{ cm}$, (ii) $5\text{ cm}$ in front of a concave mirror of radius of curvature $15\text{ cm}$. Find position, nature, and magnification.

Given: $f = -R/2 = -7.5\text{ cm}$.
(i) $u = -10\text{ cm}$: $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-7.5} - \frac{1}{-10} = -\frac{1}{30} \implies v = -30\text{ cm}$$ $$m = -\frac{v}{u} = -\frac{-30}{-10} = -3 \quad (\text{Real, inverted, magnified 3 times})$$
(ii) $u = -5\text{ cm}$: $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-7.5} - \frac{1}{-5} = +\frac{1}{15} \implies v = +15\text{ cm}$$ $$m = -\frac{v}{u} = -\frac{+15}{-5} = +3 \quad (\text{Virtual, erect, magnified 3 times})$$
Example 9.4

Convex Mirror Side-View & Image Velocity

A jogger approaches a parked car at $5\text{ m s}^{-1}$ in a convex side-view mirror with $R = 2\text{ m}$ ($f = +1\text{ m}$). How fast does the image appear to move when the jogger is at (a) $39\text{ m}$, (b) $29\text{ m}$, (c) $19\text{ m}$, (d) $9\text{ m}$?

Formula: $v = \frac{f u}{u - f}$. In $1\text{ s}$, jogger moves $5\text{ m}$ closer.
  • (a) $u = -39\text{ m} \to -34\text{ m}$: $v_1 = \frac{39}{40}\text{ m}, v_2 = \frac{34}{35}\text{ m} \implies \Delta v = \frac{1}{280}\text{ m} \implies v_{\text{img}} = \frac{1}{280}\text{ m s}^{-1}$.
  • (b) $u = -29\text{ m} \to -24\text{ m}$: $v_{\text{img}} = \frac{1}{150}\text{ m s}^{-1}$.
  • (c) $u = -19\text{ m} \to -14\text{ m}$: $v_{\text{img}} = \frac{1}{60}\text{ m s}^{-1}$.
  • (d) $u = -9\text{ m} \to -4\text{ m}$: $v_{\text{img}} = \frac{1}{10}\text{ m s}^{-1}$.
Conclusion: The apparent image speed accelerates rapidly as the object approaches the mirror.
9.3 & 9.4
NCERT Sections

Refraction & Total Internal Reflection (TIR)

Snell's Law & Refractive Index

9.3.1 Snell's Law, Apparent Depth & Lateral Shift

  • Snell's Law: $$n_{21} = \frac{n_2}{n_1} = \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}, \qquad n_1 \sin i = n_2 \sin r$$
  • Apparent Depth: When viewed near normal from rarer medium ($n_1 = 1$) into denser medium ($n_2 = n$): $$h_{\text{apparent}} = \frac{h_{\text{real}}}{n}, \qquad \Delta h = h_{\text{real}}\left(1 - \frac{1}{n}\right) \quad [\text{Normal Apparent Shift}]$$
  • Lateral Shift ($x$) in a Parallel Glass Slab of thickness $t$: $$x = \frac{t \sin(i - r)}{\cos r}$$
Total Internal Reflection

9.4.1 Critical Angle ($i_c$) & Applications (Optical Fibres & Prisms)

  • Conditions for TIR:
    1. Light must travel from an optically denser medium ($n_1$) towards a rarer medium ($n_2$).
    2. Angle of incidence must exceed critical angle: $i > i_c$.
$$\sin i_c = \frac{n_2}{n_1} \implies n = \frac{1}{\sin i_c} \quad (\text{for rarer medium = air})$$
  • Critical Angles: Water ($n=1.33 \implies i_c = 48.75^\circ$), Crown glass ($n=1.52 \implies i_c = 41.14^\circ$), Dense flint ($n=1.62 \implies i_c = 37.31^\circ$), Diamond ($n=2.42 \implies i_c = 24.41^\circ$).
  • Totally Reflecting Prisms ($45^\circ-90^\circ-45^\circ$): Bends light by $90^\circ$ (periscope), by $180^\circ$ (binoculars), or inverts image without lateral shift (Porro prism).
  • Optical Fibres: Core ($n_{\text{core}}$) coated by cladding ($n_{\text{clad}} < n_{\text{core}}$). Light enters at acceptance angle and travels by multiple TIRs with negligible attenuation ($>95\%$ transmission over $1\text{ km}$). Used as light pipes in medical endoscopes and optical communication.
Fig 9.13: Total Internal Reflection in Optical Fibre Core (n_core > n_clad)
Cladding (n_clad < n_core) Cladding (n_clad < n_core) Core (n_core) TIR (i > i_c) TIR (i > i_c)
9.5
NCERT Section

Refraction at Spherical Surfaces & Thin Lenses

Master Formulae

9.5.1 Single Spherical Interface & Lens Maker's Formula

  • Refraction at Single Spherical Surface: $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$
  • Lens Maker's Formula (Double Convex Lens in medium $n_1$): $$\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
  • Thin Lens Formula & Magnification: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \qquad m = \frac{h'}{h} = \frac{v}{u} = \frac{f - v}{f} = \frac{f}{f + u}$$
  • Power of a Lens ($P$): $$P = \frac{1}{f\text{ (in metres)}} \quad [\text{Unit: Dioptre (D)} = \text{m}^{-1}]$$
  • Coaxial Thin Lenses in Contact: $$\frac{1}{f_{\text{net}}} = \frac{1}{f_1} + \frac{1}{f_2} + \dots, \qquad P_{\text{net}} = P_1 + P_2 + \dots, \qquad m_{\text{net}} = m_1 \times m_2 \times \dots$$
Example 9.5

Refraction at Spherical Glass Interface

Light from a point source in air ($n_1 = 1$) falls on a spherical glass surface ($n_2 = 1.5$, $R = +20\text{ cm}$) at object distance $u = -100\text{ cm}$. Find image position $v$.

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \implies \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{+20} = \frac{0.5}{20} = \frac{1}{40}$$ $$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5 - 2}{200} = \frac{3}{200} \implies v = 1.5 \times \frac{200}{3} = +100\text{ cm}$$
Answer: $v = +100\text{ cm}$ (formed in glass $100\text{ cm}$ to the right).
Example 9.6

Lens Disappearance in Liquid

A glass lens with $n = 1.47$ disappears when immersed in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?

Condition: For lens to disappear, no refraction must occur at its surface $\implies n_{\text{liquid}} = n_{\text{lens}} = 1.47$. Then $\frac{1}{f} = \left(\frac{1.47}{1.47} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0 \implies f \to \infty$.
Liquid Identity: No, it cannot be water ($n = 1.33$). It is typically glycerine or a specialized oil mixture.
Example 9.7

Lens Power, Radii & Focal Length in Liquid

(i) If $f = +0.5\text{ m}$, what is $P$?
(ii) Double convex lens has $R_1 = +10\text{ cm}, R_2 = -15\text{ cm}, f = +12\text{ cm}$. Find $n$.
(iii) Convex lens has $f_{\text{air}} = 20\text{ cm}$. Find $f_{\text{water}}$ ($n_g = 1.5, n_w = 1.33$).

(i) Power: $P = \frac{1}{f} = \frac{1}{0.5\text{ m}} = +2.0\text{ D}$.
(ii) Refractive Index: $$\frac{1}{12} = (n - 1)\left(\frac{1}{10} - \frac{1}{-15}\right) = (n - 1)\left(\frac{1}{6}\right) \implies n - 1 = \frac{6}{12} = 0.5 \implies n = 1.5$$
(iii) Focal Length in Water: $$\frac{1}{f_w} = \left(\frac{n_g}{n_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1.5}{1.333} - 1\right)\frac{1}{f_a (n_g - 1)} = \frac{0.125}{0.5 \times 20} \implies f_w = +78.2\text{ cm}$$
Example 9.8

Successive Image Formation by 3-Lens System

Find the final image position for 3 coaxial lenses: Convex $f_1 = +10\text{ cm}$ ($u = -30\text{ cm}$), Concave $f_2 = -10\text{ cm}$ ($d_1 = 5\text{ cm}$), Convex $f_3 = +30\text{ cm}$ ($d_2 = 10\text{ cm}$).

Lens 1: $\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \implies v_1 = +15\text{ cm}$.
Lens 2: Object at $u_2 = 15 - 5 = +10\text{ cm}$. $$\frac{1}{v_2} - \frac{1}{+10} = \frac{1}{-10} \implies \frac{1}{v_2} = 0 \implies v_2 = \infty$$
Lens 3: Parallel rays incident ($u_3 = -\infty$) $\implies v_3 = f_3 = +30\text{ cm}$.
Final Image is formed $30\text{ cm}$ to the right of the third lens.
9.6
NCERT Section

Refraction Through a Prism

Prism Formula & Minimum Deviation

9.6.1 Deviation & Refractive Index Formula

  • Prism Geometry: For angle of prism $A$, incident angle $i$, and emergent angle $e$: $$r_1 + r_2 = A, \qquad \delta = i + e - A$$
  • Condition of Minimum Deviation ($D_m$):
    Inside the prism, ray travels symmetrically parallel to base: $$i = e \implies r_1 = r_2 = r = \frac{A}{2}, \qquad D_m = 2i - A \implies i = \frac{A + D_m}{2}$$
$$n_{21} = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \quad [\text{Prism Formula}], \qquad D_m \approx (n - 1) A \quad [\text{Thin Prism}]$$
Fig 9.24: Refraction Through Prism & Minimum Deviation (Dm)
A B C δ (Deviation) i e r₁ = r₂ (At Dm) n = sin[(A+Dm)/2] / sin(A/2)
9.7
NCERT Section

Optical Instruments: Microscopes & Telescopes

Microscopes

9.7.1 Simple & Compound Microscope Magnifying Power

  • Simple Microscope (Single Converging Lens):
    • Image at Near Point ($D = 25\text{ cm}$): $m = 1 + \frac{D}{f}$
    • Image at Infinity (Relaxed Eye / Normal Adjustment): $m = \frac{D}{f}$
  • Compound Microscope (Objective $f_o$ + Eyepiece $f_e$):
    Objective produces real, inverted, magnified image ($m_o = \frac{v_o}{u_o} \approx \frac{L}{f_o}$), eyepiece acts as simple magnifier.
    • Final Image at Infinity: $$m = m_o \times m_e = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right), \qquad L_{\text{tube}} = v_o + f_e \approx L + f_o + f_e$$
    • Final Image at Near Point ($D$): $$m = \left(\frac{L}{f_o}\right)\left(1 + \frac{D}{f_e}\right), \qquad L_{\text{tube}} = v_o + |u_e|$$
    For high magnification, both $f_o$ and $f_e$ must be very small.
Astronomical & Reflecting Telescopes

9.7.2 Refracting & Reflecting (Cassegrain) Telescopes

  • Astronomical Refracting Telescope:
    Objective has large focal length $f_o$ and large aperture; eyepiece has short focal length $f_e$.
    • Normal Adjustment (Image at Infinity): $$m = \frac{f_o}{f_e}, \qquad L = f_o + f_e$$
    • Image at Near Point ($D$): $$m = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right), \qquad L = f_o + u_e$$
  • Reflecting (Cassegrain) Telescope Advantages:
    1. No chromatic aberration (reflection is independent of wavelength $\lambda$).
    2. Spherical aberration is eliminated by using parabolic primary mirrors.
    3. Much lighter and mechanically easier to support from the rear.
Fig 9.29: Astronomical Refracting Telescope Ray Diagram (Normal Adjustment)
Objective (f_o, Large) Eyepiece (f_e) I' (at F_o / F_e) To Infinity (Relaxed Eye) Normal Adjustment Magnification: m = f_o / f_e • Tube Length L = f_o + f_e
02 / Interactive Assessment

Concept Check & Board Mastery Quiz

15 targeted MCQs covering spherical mirrors, sign conventions, refraction, total internal reflection, lens maker's formula, prisms, and optical instruments with in-depth explanations.

15 MCQS
0 / 15 Answered
Score: 0 (0%)
03 / Textbook Solutions

NCERT Exercises 9.1 – 9.31

Complete stepwise solutions for every single textbook exercise in NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Reprint 2026-27.

31 EXERCISES
Ex 9.1Concave Mirror Real Image

9.1 A candle $2.5\text{ cm}$ in size is placed at $27\text{ cm}$ in front of a concave mirror of radius $36\text{ cm}$. Find screen distance, image size, and nature. How should the screen be moved if the candle is moved closer?

$v = -54\text{ cm}$, Image size $h' = -5.0\text{ cm}$ (Real, Inverted) • Screen moved away from mirror.
$u = -27\text{ cm}, f = -18\text{ cm}$. $\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{54} \implies v = -54\text{ cm}$.
$m = -\frac{v}{u} = -\frac{-54}{-27} = -2 \implies h' = -2 \times 2.5\text{ cm} = -5.0\text{ cm}$.
As candle moves closer to focus, $v$ increases and screen must be moved farther away. If $u < f$, virtual image forms behind mirror (cannot be caught on screen).
Ex 9.2Convex Mirror Image

9.2 A $4.5\text{ cm}$ needle is placed $12\text{ cm}$ from a convex mirror of focal length $15\text{ cm}$. Find location and magnification. What happens as needle moves farther?

$v = +6.7\text{ cm}$ (behind mirror), $m = +0.556$, $h' = 2.5\text{ cm}$ (Virtual, Erect).
$u = -12\text{ cm}, f = +15\text{ cm} \implies \frac{1}{v} = \frac{1}{15} - \frac{1}{-12} = \frac{9}{60} \implies v = \frac{20}{3} = +6.67\text{ cm}$.
$m = -\frac{v}{u} = -\frac{20/3}{-12} = +\frac{5}{9} \approx +0.56 \implies h' = 2.5\text{ cm}$.
As needle moves farther ($u \to -\infty$), image moves toward focus $F$ ($v \to +15\text{ cm}$) and size decreases toward zero.
Ex 9.3Apparent Depth & Refractive Index

9.3 A tank has water to height $12.5\text{ cm}$. Apparent depth is $9.4\text{ cm}$. Find $n_{\text{water}}$. If replaced by liquid of $n = 1.63$, by what distance is the microscope moved?

$n_{\text{water}} = 1.33$ • Shift = $1.73\text{ cm}$ downwards.
$n_w = \frac{h_{\text{real}}}{h_{\text{app}}} = \frac{12.5\text{ cm}}{9.4\text{ cm}} \approx 1.33$.
With liquid $n = 1.63$: $h'_{\text{app}} = \frac{12.5\text{ cm}}{1.63} = 7.67\text{ cm}$.
Microscope must be moved down by $\Delta y = 9.4 - 7.67 = 1.73\text{ cm}$.
Ex 9.4Snell's Law at Multiple Interfaces

9.4 For air-glass $i = 60^\circ, r = 35^\circ$; for air-water $i = 60^\circ, r = 47^\circ$. Predict angle of refraction in glass for water-glass interface at $i = 45^\circ$.

$r = 38.2^\circ$ in glass.
$n_g = \frac{\sin 60^\circ}{\sin 35^\circ} = \frac{0.8660}{0.5736} = 1.51$. $n_w = \frac{\sin 60^\circ}{\sin 47^\circ} = \frac{0.8660}{0.7314} = 1.184 \implies {}^w n_g = \frac{n_g}{n_w} = \frac{1.51}{1.184} = 1.275$.
Snell's law: $\sin r = \frac{\sin 45^\circ}{{}^w n_g} = \frac{0.7071}{1.275} = 0.5546 \implies r = 38.2^\circ$.
Ex 9.5Circle of Illuminance (TIR Cone)

9.5 A bulb is at the bottom of water ($n = 1.33$) of depth $80\text{ cm}$. Find the surface area through which light emerges.

Area $A = 2.58\text{ m}^2$.
Critical angle: $\sin i_c = 1/n = 1/1.33 = 0.7519 \implies \tan i_c = \frac{1}{\sqrt{n^2 - 1}} = \frac{1}{\sqrt{1.33^2 - 1}} = 1.136$.
Radius of circle of illumination: $r = d \tan i_c = \frac{0.80}{\sqrt{1.33^2 - 1}} = \frac{0.80}{0.876} = 0.907\text{ m}$.
$$A = \pi r^2 = \pi \frac{d^2}{n^2 - 1} = \frac{3.1416 \times 0.64}{0.7689} = 2.58\text{ m}^2$$
Ex 9.6Prism Formula in Air and Water

9.6 A $60^\circ$ prism has $D_m = 40^\circ$ in air. Find $n_{\text{prism}}$. If immersed in water ($n=1.33$), find new $D'_m$.

$n_{\text{prism}} = 1.532$ • New minimum deviation in water $D'_m = 10.32^\circ$.
$n = \frac{\sin((60^\circ + 40^\circ)/2)}{\sin(60^\circ/2)} = \frac{\sin 50^\circ}{\sin 30^\circ} = \frac{0.7660}{0.5} = 1.532$.
In water: ${}^w n_g = \frac{1.532}{1.333} = 1.149 \implies \sin\left(\frac{60^\circ + D'_m}{2}\right) = 1.149 \times \sin 30^\circ = 0.5746$.
$\frac{60^\circ + D'_m}{2} = 35.16^\circ \implies D'_m = 70.32^\circ - 60^\circ = 10.32^\circ$.
Ex 9.7Equiconvex Lens Maker's Formula

9.7 Double-convex lens of $n = 1.55$ has both faces of equal radius $R$. Find $R$ for $f = 20\text{ cm}$.

Radius of curvature $R = 22.0\text{ cm}$.
$\frac{1}{f} = (n - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = \frac{2(n - 1)}{R} \implies R = 2(n - 1) f = 2(1.55 - 1) \times 20\text{ cm} = 22.0\text{ cm}$.
Ex 9.8Convergent Beam on Lenses

9.8 A beam converges at $P$. Lens is placed $12\text{ cm}$ from $P$ ($u = +12\text{ cm}$). Where does it converge for (a) convex $f = 20\text{ cm}$, (b) concave $f = -16\text{ cm}$?

(a) $v = +7.5\text{ cm}$ • (b) $v = +48\text{ cm}$.
(a) Convex: $\frac{1}{v} - \frac{1}{+12} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{8}{60} \implies v = +7.5\text{ cm}$.
(b) Concave: $\frac{1}{v} - \frac{1}{+12} = \frac{1}{-16} \implies \frac{1}{v} = -\frac{1}{16} + \frac{1}{12} = \frac{1}{48} \implies v = +48\text{ cm}$.
Ex 9.9Concave Lens Image

9.9 Object $3.0\text{ cm}$ is placed $14\text{ cm}$ in front of concave lens $f = -21\text{ cm}$. Describe image and effect of moving object farther.

$v = -8.4\text{ cm}$, $h' = +1.8\text{ cm}$ (Virtual, Erect, Diminished).
$\frac{1}{v} = \frac{1}{-21} + \frac{1}{-14} = -\frac{5}{42} \implies v = -8.4\text{ cm}$.
$m = \frac{v}{u} = \frac{-8.4}{-14} = +0.6 \implies h' = 0.6 \times 3.0 = 1.8\text{ cm}$. As $u \to -\infty$, image shifts towards focus ($v \to -21\text{ cm}$) and size shrinks to zero.
Ex 9.10Combination of Lenses in Contact

9.10 Convex lens $f_1 = +30\text{ cm}$ is in contact with concave lens $f_2 = -20\text{ cm}$. Find effective focal length and nature.

$F = -60\text{ cm}$ (Diverging system).
$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = -\frac{1}{60} \implies F = -60\text{ cm}$. Power $P = -\frac{100}{60} = -1.67\text{ D}$.
Ex 9.11Compound Microscope Calculations

9.11 Compound microscope has $f_o = 2.0\text{ cm}, f_e = 6.25\text{ cm}$, separation $15\text{ cm}$. Find object distance and magnifying power for (a) final image at $25\text{ cm}$, (b) final image at infinity.

(a) $u_o = -2.59\text{ cm}, m = 20$ • (b) $u_o = -2.46\text{ cm}, m = 13.5$.
(a) Image at $D = 25\text{ cm}$: $\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25} \implies u_e = -5.0\text{ cm}$.
$v_o = 15 - 5 = 10\text{ cm} \implies \frac{1}{10} - \frac{1}{u_o} = \frac{1}{2} \implies u_o = -2.5\text{ cm}$.
$m = \left(\frac{v_o}{|u_o|}\right)\left(1 + \frac{D}{f_e}\right) = \left(\frac{10}{2.5}\right)(1 + 4) = 4 \times 5 = 20$.
(b) Image at $\infty$: $u_e = -f_e = -6.25\text{ cm} \implies v_o = 15 - 6.25 = 8.75\text{ cm}$.
$\frac{1}{8.75} - \frac{1}{u_o} = \frac{1}{2} \implies u_o = -2.59\text{ cm}$.
$m = \left(\frac{v_o}{|u_o|}\right)\left(\frac{D}{f_e}\right) = \left(\frac{8.75}{2.59}\right)\left(\frac{25}{6.25}\right) = 3.38 \times 4 = 13.5$.
Ex 9.12Microscope Tube Length & Magnification

9.12 $f_o = 8.0\text{ mm}, f_e = 2.5\text{ cm}$, object at $u_o = -9.0\text{ mm}$, image at $D = 25\text{ cm}$. Find lens separation and magnifying power.

Separation $L = 9.47\text{ cm}$, Magnifying Power $m = 88$.
$\frac{1}{v_o} - \frac{1}{-0.9} = \frac{1}{0.8} \implies v_o = 7.2\text{ cm}$.
$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{2.5} \implies u_e = -2.27\text{ cm}$.
Separation $L = v_o + |u_e| = 7.2 + 2.27 = 9.47\text{ cm}$.
$m = \left(\frac{v_o}{|u_o|}\right)\left(1 + \frac{D}{f_e}\right) = \left(\frac{7.2}{0.9}\right)\left(1 + \frac{25}{2.5}\right) = 8 \times 11 = 88$.
Ex 9.13Astronomical Telescope Normal Adjustment

9.13 Telescope has $f_o = 144\text{ cm}, f_e = 6.0\text{ cm}$. Find magnifying power and separation in normal adjustment.

$m = 24$, Separation $L = 150\text{ cm}$.
$m = \frac{f_o}{f_e} = \frac{144}{6} = 24$. Separation $L = f_o + f_e = 144 + 6 = 150\text{ cm}$.
Ex 9.14Giant Telescope Moon Viewing

9.14 (a) Giant telescope has $f_o = 15\text{ m}, f_e = 1.0\text{ cm}$. Find angular magnification. (b) Find diameter of moon's image formed by objective ($d_{\text{moon}} = 3.48 \times 10^6\text{ m}, r = 3.8 \times 10^8\text{ m}$).

(a) $m = 1500$ • (b) Image diameter $d = 13.73\text{ cm}$.
(a) $m = \frac{f_o}{f_e} = \frac{15\text{ m}}{0.01\text{ m}} = 1500$.
(b) Angle subtended by moon $\alpha = \frac{3.48 \times 10^6\text{ m}}{3.8 \times 10^8\text{ m}} = 9.158 \times 10^{-3}\text{ rad}$.
Diameter of image at objective focus $d = f_o \alpha = 15\text{ m} \times (9.158 \times 10^{-3}) = 0.1374\text{ m} = 13.74\text{ cm}$.
Ex 9.15Algebraic Deduction from Mirror Formula

9.15 Use mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ to algebraically deduce: (a) Object between $f$ and $2f$ of concave mirror gives real image beyond $2f$. (b) Convex mirror always gives virtual image. (c) Convex mirror image is always diminished and between pole and focus. (d) Object between pole and focus of concave mirror gives virtual enlarged image.

(a) Concave ($f < 0$), $-2f < u < -f$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$. Since $2f < u < f \implies \frac{1}{2f} > \frac{1}{u} > \frac{1}{f} \implies \frac{1}{v} < 0$ and $v < 2f$ (Real, formed beyond $2f$).
(b) Convex ($f > 0$), $u < 0$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{|u|} > 0 \implies v > 0$ (Always virtual).
(c) Convex mirror: $\frac{1}{v} > \frac{1}{f} \implies v < f$, so $0 < v < f$. Magnification $m = \frac{v}{|u|} < 1$ (Always diminished).
(d) Concave ($f < 0$), $|u| < |f|$: $\frac{1}{v} = \frac{1}{f} + \frac{1}{|u|} > 0 \implies v > 0$ (Virtual) and $m = \frac{v}{|u|} > 1$ (Enlarged).
Ex 9.16Apparent Shift by Glass Slab

9.16 Pin on table viewed through $15\text{ cm}$ glass slab ($n = 1.5$). By what distance does it appear raised? Does it depend on slab location?

Apparent Shift $\Delta t = 5.0\text{ cm}$ • Independent of slab position.
$\Delta t = t\left(1 - \frac{1}{n}\right) = 15\left(1 - \frac{1}{1.5}\right) = 15 \times \frac{1}{3} = 5.0\text{ cm}$. The formula is completely independent of where the slab is placed between the object and observer.
Ex 9.17Acceptance Angle of Optical Fibre

9.17 Light pipe core $n_1 = 1.68$, cladding $n_2 = 1.44$. (a) Find maximum incident angle $i$ with axis for TIR. (b) What if there is no cladding ($n_2 = 1$)?

(a) $0 \le i \le 60^\circ$ • (b) All rays $0 \le i \le 90^\circ$ undergo TIR.
(a) $\sin i_c = \frac{1.44}{1.68} = 0.8571 \implies i_c = 59^\circ$. Refraction angle $r = 90^\circ - 59^\circ = 31^\circ$.
$\sin i_{\text{max}} = n_1 \sin r = 1.68 \times \sin 31^\circ = 1.68 \times 0.515 = 0.865 \implies i_{\text{max}} = 60^\circ$.
(b) With air: $\sin i_c = 1/1.68 = 0.5952 \implies i_c = 36.5^\circ \implies r = 53.5^\circ$. $\sin i = 1.68 \sin 53.5^\circ = 1.35 > 1$, meaning all rays incident from $0^\circ$ to $90^\circ$ undergo total internal reflection.
Ex 9.18Maximum Focal Length for Real Image on Wall

9.18 Real image of bulb on opposite wall $D = 3\text{ m}$ away. What is maximum focal length $f_{\text{max}}$?

$f_{\text{max}} = 0.75\text{ m} = 75\text{ cm}$.
For a real image on a screen separated by distance $D$, the condition is $D \ge 4f \implies f \le D/4 = 3.0/4 = 0.75\text{ m}$.
Ex 9.19Displacement Method for Focal Length

9.19 Screen is $90\text{ cm}$ from object ($D = 90\text{ cm}$). Two lens positions separated by $d = 20\text{ cm}$ form sharp images. Find $f$.

$f = 21.4\text{ cm}$.
Displacement formula: $f = \frac{D^2 - d^2}{4D} = \frac{90^2 - 20^2}{4 \times 90} = \frac{8100 - 400}{360} = \frac{7700}{360} = 21.39\text{ cm} \approx 21.4\text{ cm}$.
Ex 9.20Separated Lenses System

9.20 Lenses $f_1 = +30\text{ cm}, f_2 = -20\text{ cm}$ separated by $8.0\text{ cm}$. (a) Effective focal length for parallel light from both sides. (b) Image size for $1.5\text{ cm}$ object at $40\text{ cm}$ from convex lens.

(a) $f = -216\text{ cm}$ (effective concept not useful as focus depends on side) • (b) Image at $420\text{ cm}$ from concave lens, size $h' = 0.98\text{ cm}$.
Convex lens produces image at $v_1 = 120\text{ cm}$. Concave lens receives virtual object at $u_2 = 120 - 8 = +112\text{ cm} \implies v_2 = -420\text{ cm}$. Net magnification $m = m_1 \times m_2 = (-3) \times (+3.75) = -11.25 \implies h' = 0.98\text{ cm}$.
Ex 9.21Grazing Emergence in Prism

9.21 Prism $A = 60^\circ, n = 1.524$. Find incident angle $i$ for grazing emergence ($r_2 = i_c$).

Incident angle $i = 29.75^\circ \approx 30^\circ$.
$\sin i_c = 1/1.524 = 0.6562 \implies i_c = 41.0^\circ = r_2$.
$r_1 = A - r_2 = 60^\circ - 41.0^\circ = 19.0^\circ$.
$\sin i = n \sin r_1 = 1.524 \times \sin 19.0^\circ = 1.524 \times 0.3256 = 0.4962 \implies i = 29.75^\circ$.
Ex 9.22 to 9.24Magnification vs Magnifying Power

9.22–9.24 Magnifying glass analysis ($f = 9\text{ cm}$, object at $9\text{ cm}$ and $D = 25\text{ cm}$).

9.22: Object at $u = -9\text{ cm} = -f \implies v \to \infty$. Linear magnification $m = v/u \to \infty$. Angular magnifying power $m_{\text{ang}} = D/f = 25/9 = 2.78$. Linear magnification is infinite because image is at infinity, but angular size remains finite.
9.23: For maximum angular magnifying power, image forms at $D = 25\text{ cm}$: $u = -6.62\text{ cm}$. $m_{\text{max}} = 1 + D/f = 1 + 2.78 = 3.78$. Here linear magnification $m = v/u = 25/6.62 = 3.78 = m_{\text{ang}}$.
9.24: Virtual image area $6.25\text{ mm}^2 \implies$ linear magnification $m = \sqrt{6.25} = 2.5$. $v = -2.5 u \implies u = -5.4\text{ cm}, v = -13.5\text{ cm}$. Since $v = 13.5\text{ cm} < 25\text{ cm}$ (less than near point), the eye cannot focus distinctly on it.
Ex 9.25Conceptual Optics Questions

9.25 Detailed conceptual answers on simple & compound microscopes.

(a) A magnifier allows the object to be brought closer to the eye ($u < 25\text{ cm}$) than the unassisted near point ($25\text{ cm}$), thereby subtending a larger visual angle.
(b) Moving eye back reduces angular magnification slightly because field of view decreases.
(c) Very small $f$ requires highly curved surfaces, causing severe spherical and chromatic aberrations.
(d) Both $f_o$ and $f_e$ must be small to maximize magnification ($m \propto 1/(f_o f_e)$).
(e) Eye should be placed at the eye-ring (image of objective formed by eyepiece) to collect all refracted rays with maximum field of view.
Ex 9.26 to 9.28Microscope Setup & Telescope Tower Viewing

9.26–9.28 Compound microscope $30\text{X}$ setup and telescope viewing a $100\text{ m}$ tall tower $3\text{ km}$ away.

9.26: $f_o = 1.25\text{ cm}, f_e = 5.0\text{ cm}, m = 30$. With image at $D = 25\text{ cm}$, $m_e = 1 + 25/5 = 6 \implies m_o = 30/6 = 5$. $v_o = 7.5\text{ cm}, u_o = -1.5\text{ cm}$. Tube separation $L = v_o + |u_e| = 7.5 + 4.17 = 11.67\text{ cm}$.
9.27: $f_o = 140\text{ cm}, f_e = 5\text{ cm}$. (a) $m = 140/5 = 28$. (b) $m = \frac{f_o}{f_e}(1 + f_e/D) = 28(1 + 0.2) = 33.6$.
9.28: (a) $L = 145\text{ cm}$. (b) $\theta = \frac{100}{3000} = \frac{1}{30}\text{ rad} \implies h_1 = 140 \times \frac{1}{30} = 4.67\text{ cm}$. (c) Final image height $h_2 = m_e h_1 = \left(1 + \frac{25}{5}\right) \times 4.67 = 6 \times 4.67 = 28.0\text{ cm}$.
Ex 9.29 to 9.31Cassegrain Telescope & Liquid Lens Refractive Index

9.29 Cassegrain telescope ($R_1 = 220\text{ mm}, R_2 = 140\text{ mm}, d = 20\text{ mm}$).
9.30 Galvanometer mirror optical lever ($3.5^\circ$ deflection, $D = 1.5\text{ m}$).
9.31 Equiconvex lens ($n = 1.50$) on liquid layer and mirror ($f_1 = 30\text{ cm}, F = 45\text{ cm}$).

9.29 Cassegrain: Primary concave mirror ($f_1 = 110\text{ mm}$) forms image at $110\text{ mm}$. Secondary convex mirror ($f_2 = 70\text{ mm}$) at $d = 20\text{ mm}$ receives virtual object at $u = 110 - 20 = +90\text{ mm}$. $\frac{1}{v} + \frac{1}{90} = \frac{1}{70} \implies v = +315\text{ mm} = 31.5\text{ cm}$ behind secondary mirror.
9.30 Galvanometer: Mirror deflection $\theta = 3.5^\circ \implies$ reflected ray rotates by $2\theta = 7.0^\circ$. Displacement $d = D \tan(7.0^\circ) = 1.5 \times 0.1228 = 0.184\text{ m} = 18.4\text{ cm}$.
9.31 Liquid Layer: Glass lens $f_1 = 30\text{ cm} \implies R = 30\text{ cm}$. Combined focal length $F = 45\text{ cm} \implies \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \implies \frac{1}{f_2} = \frac{1}{45} - \frac{1}{30} = -\frac{1}{90} \implies f_2 = -90\text{ cm}$ (plano-concave liquid lens).
$\frac{1}{f_2} = (n_l - 1)\left(-\frac{1}{R}\right) \implies -\frac{1}{90} = (n_l - 1)\left(-\frac{1}{30}\right) \implies n_l - 1 = \frac{30}{90} = 0.333 \implies n_{\text{liquid}} = 1.33$.
04 / Rapid Reference

Chapter Summary & Points to Ponder

Core ray optics formulae, sign conventions, optical instrument magnification summaries, and official NCERT Points to Ponder for Chapter 9.

100% SYLLABUS

Mirror Formula & Magnification

1/v + 1/u = 1/f • m = −v/u

Distances along incident light are positive; focal length $f = R/2$ (negative for concave, positive for convex).

Refraction at Spherical Surface

n₂/v − n₁/u = (n₂ − n₁)/R

Master equation relating object and image distances across a single curved interface separating two media.

Lens Maker's & Thin Lens Formula

1/f = (n−1)(1/R₁ − 1/R₂) • 1/v − 1/u = 1/f

Lens power $P = 1/f$ (Dioptres). In contact: $P_{\text{net}} = \sum P_i$ and $m_{\text{net}} = \prod m_i$.

Total Internal Reflection (TIR)

sin i_c = 1/n • i > i_c

Requires light travelling from denser to rarer medium. Underpins optical fibres, prisms, and mirages.

Prism Formula

n = sin((A + D_m)/2) / sin(A/2)

At minimum deviation $D_m$, ray travels parallel to base ($r = A/2$). For thin prism: $D_m \approx (n-1)A$.

Microscope & Telescope Magnification

m_{\mu} = (L/f_o)(D/f_e) • m_{\text{tel}} = f_o / f_e

Microscope requires small $f_o, f_e$; Telescope requires large $f_o$ and large aperture for resolving power.

NCERT Official

Points to Ponder

  1. Universality of Reflection & Refraction Laws: The laws $\theta_i = \theta_r$ and $n_1 \sin i = n_2 \sin r$ apply strictly at every single microscopic point of incidence on plane or curved interfaces.
  2. Existence of Real Image without a Screen: A real image exists suspended in physical space where rays actually intersect, whether or not a screen is placed to diffuse the light into our eyes.
  3. Regular vs Diffuse Reflection: Regular specular reflection brings all rays from an object point to a unique image point; irregular scattering off paper or walls prevents image formation.
  4. Dispersion and Lens Chromatic Aberration: Thick lenses show coloured fringes because refractive index $n(\lambda)$ varies with wavelength ($n_{\text{violet}} > n_{\text{red}}$). Mirrors suffer zero chromatic aberration.
  5. Microscope Visual Angle Advantage: The magnifying glass allows an object to be brought much closer to the eye ($u \ll 25\text{ cm}$) than the unassisted near point ($25\text{ cm}$), subtending a much larger visual angle on the retina.
05 / Practice Tests

3-Tier Practice Tests

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