∠i = ∠r (angle of incidence = angle of reflection). The incident ray, reflected ray and normal are coplanar. Valid for all reflecting surfaces.
Convention
➕➖ Cartesian Sign Convention
All distances from pole P. Incident light direction = positive. Heights above principal axis = positive. Heights below = negative.
Key Formula
🔍 Focal Length
For a spherical mirror: f = R/2, where R = radius of curvature. f is negative for concave, positive for convex.
Mirror Equation
1/v + 1/u = 1/f | f = R/2
Magnification
m = h'/h = −v/u
where u = object distance, v = image distance, f = focal length
🖼️[NCERT Fig. 9.1: Incident ray, reflected ray & normal — coplanar]
Add the concave/convex mirror ray diagrams here (NCERT Fig. 9.3, 9.5, 9.6)
Concave Mirror
🔆 Image Cases — Concave
Beyond C: Real, inverted, diminished between F&C At C: Real, inverted, same size Between F&C: Real, inverted, magnified At F: At infinity Between P&F: Virtual, erect, magnified
Convex Mirror
🔆 Image Cases — Convex
For any position of object: Image is always virtual, erect and diminished. Image is located between Pole (P) and Focus (F). Used in rear-view mirrors.
🎯 Practice MCQs — Section 9.2
Q4. A concave mirror has radius of curvature 30 cm. Its focal length is: Easy
A) +30 cm
B) −15 cm
C) +15 cm
D) −30 cm
f = R/2 = 30/2 = 15 cm. For concave mirror f is negative → f = −15 cm. Option B is correct.
Q5. An object is placed 10 cm in front of a concave mirror of f = −7.5 cm. The image distance v is: Moderate
A) −15 cm
B) +30 cm
C) −30 cm
D) +15 cm
1/v + 1/u = 1/f → 1/v + 1/(−10) = 1/(−7.5) → 1/v = −1/7.5 + 1/10 = (−10+7.5)/75 = −2.5/75 = −1/30 → v = −30 cm. Real image, 30 cm in front. Option C.
Q6. A convex mirror of f = +10 cm has an object at 30 cm in front. The magnification is: Hard
A) −0.25
B) +0.25
C) +0.5
D) −0.5
u = −30, f = +10. 1/v = 1/f − 1/u = 1/10 − 1/(−30) = 1/10 + 1/30 = 4/30 → v = +7.5 cm. m = −v/u = −7.5/(−30) = +0.25. Virtual, erect, diminished. Option B.
💧
9.3 Refraction of Light
Snell's Law | Refractive index | Lateral shift | Apparent depth
Snell's Law
📏 Law of Refraction
At interface of two media: n₁ sin i = n₂ sin r. The ratio sin i / sin r = n₂₁ = constant (refractive index of medium 2 w.r.t medium 1).
Refractive Index
🔢 Key Relations
n = c/v (speed ratio)
n₁₂ = 1/n₂₁
n₃₂ = n₃₁ × n₁₂
Higher n → optically denser medium → light bends toward normal.
Application
🐟 Apparent Depth
Object in denser medium appears closer: Apparent depth = Real depth / n e.g. pool bottom looks raised because water has n = 1.33.
Snell's Law
n₁ sin i = n₂ sin r | n₂₁ = sin i / sin r = c/v = n₂/n₁
Apparent Depth
h_apparent = h_real / n (for near-normal viewing)
Lateral Shift (glass slab)
Emergent ray is parallel to incident ray; no deviation but lateral shift occurs.
🖼️[NCERT Fig. 9.8: Refraction & reflection at interface | Fig. 9.9: Lateral shift | Fig. 9.10: Apparent depth]
Place refraction diagrams here
🎯 Practice MCQs — Section 9.3
Q7. A ray in air (n=1) hits glass (n=1.5) at i=30°. The angle of refraction is: Moderate
A) 30°
B) 19.47°
C) 45°
D) 12°
n₁ sin i = n₂ sin r → 1 × sin30° = 1.5 × sin r → sin r = 0.5/1.5 = 0.333 → r = sin⁻¹(0.333) ≈ 19.47°. Option B.
Q8. A tank filled with water (n=1.33) has a needle at depth 12.5 cm. The apparent depth seen from above is approximately: Moderate
A) 12.5 cm
B) 9.4 cm
C) 16.6 cm
D) 8.0 cm
Apparent depth = Real depth / n = 12.5 / 1.33 ≈ 9.4 cm. Option B. (This is NCERT Exercise 9.3)
Q9. Speed of light in a medium is 2×10⁸ m/s. The refractive index of the medium is: Easy
A) 2.0
B) 1.0
C) 1.5
D) 0.67
n = c/v = (3×10⁸)/(2×10⁸) = 1.5. Option C.
💎
9.4 Total Internal Reflection (TIR)
Critical angle | Optical fibre | Diamond | Prisms
Condition
🔁 When does TIR occur?
TIR occurs when: (1) Light travels from denser → rarer medium, AND (2) angle of incidence exceeds the critical angle (i > iᶜ).
Critical Angle
📐 Critical Angle Formula
sin iᶜ = n₂₁ = n_rarer/n_denser or n₁₂ = 1/sin iᶜ
Water: iᶜ ≈ 48.75° | Diamond: iᶜ ≈ 24.41°
Application
🌐 Optical Fibre
Core (high n) + Cladding (low n). Light undergoes repeated TIR along the fibre. Used in telecommunications, medical endoscopy. 95%+ light transmitted over 1 km.
Q11. An optical fibre uses glass core (n = 1.68) and cladding (n = 1.44). The critical angle at core-cladding interface is: Hard
A) 58.97°
B) 58.97° (≈59°)
C) 45°
D) 30°
sin iᶜ = n_cladding/n_core = 1.44/1.68 = 0.857 → iᶜ = sin⁻¹(0.857) ≈ 58.97°. (NCERT Ex 9.17). Option B.
Q12. Which of the following is NOT a phenomenon based on Total Internal Reflection? Easy
A) Optical fibre communication
B) Sparkling of diamond
C) Lateral shift of light in glass slab
D) Mirage formation
Lateral shift in a glass slab is due to refraction (not TIR). The other three phenomena — optical fibre, diamond brilliance and mirage — all involve TIR. Option C.
🔬
9.5 Refraction at Spherical Surfaces & Lenses
Lens formula | Lens maker's formula | Power | Combination
Spherical Interface
🔵 Refraction Formula
For single spherical surface between media n₁ and n₂: n₂/v − n₁/u = (n₂−n₁)/R All distances from optical centre using Cartesian convention.
Thin Lens
📡 Lens Formula
1/v − 1/u = 1/f Valid for both convex (f>0) and concave (f<0) lenses, and for both real and virtual images.
Power
⚡ Power of Lens
P = 1/f (in metres) Unit: Dioptre (D) = m⁻¹ Convex: P positive | Concave: P negative Combined: P = P₁ + P₂ + ...
🖼️[NCERT Fig. 9.16: Refraction by double convex lens | Fig. 9.17: Ray tracing convex/concave | Fig. 9.18: Power of lens]
Add lens diagrams here
🎯 Practice MCQs — Section 9.5
Q13. A convex lens of focal length 0.5 m has power equal to: Easy
A) 0.5 D
B) +2 D
C) −2 D
D) +5 D
P = 1/f = 1/0.5 = +2 D. Positive because it is a convex (converging) lens. Option B.
Q14. A double-convex lens (n=1.5) has both faces of R=20 cm. Its focal length is: Moderate
A) 20 cm
B) 10 cm
C) 40 cm
D) 15 cm
1/f = (n−1)(1/R₁−1/R₂) = (1.5−1)(1/20−1/(−20)) = 0.5×(2/20) = 0.5×0.1 = 0.05 → f = 20 cm. Option A.
Q15. A convex lens (+30 cm) in contact with concave lens (−20 cm). Net focal length of combination is: Hard
A) +50 cm
B) +10 cm
C) +25 cm
D) −60 cm
1/f = 1/f₁ + 1/f₂ = 1/30 + 1/(−20) = 1/30 − 1/20 = (2−3)/60 = −1/60 → f = −60 cm. The combination acts as a diverging lens. Option D. (NCERT Ex 9.10)
🔺
9.6 Refraction Through a Prism
Angle of deviation | Minimum deviation | Refractive index
Geometry
🔺 Prism Relations
For prism angle A: r₁ + r₂ = A δ = i + e − A where i = angle of incidence, e = angle of emergence, δ = deviation angle.
Min. Deviation
📉 At Minimum Deviation Dₘ
At Dₘ: i = e and r₁ = r₂ = A/2 r = A/2 i = (A + Dₘ)/2 Refracted ray inside prism is parallel to base.
Formula
🔢 Refractive Index of Prism
n₂₁ = sin[(A+Dₘ)/2] / sin(A/2) This allows experimental determination of n by measuring A and Dₘ.
Key Prism Equations
r₁ + r₂ = A | δ = i + e − A
At Minimum Deviation: r = A/2 | i = (A + Dₘ)/2
n₂₁ = sin[(A + Dₘ)/2] / sin(A/2)
Thin prism: Dₘ = (n₂₁ − 1)A
🖼️[NCERT Fig. 9.21: Ray through triangular prism | Fig. 9.22: δ vs i graph showing minimum deviation]
Add prism diagrams here
🎯 Practice MCQs — Section 9.6
Q16. For a prism with A = 60° and minimum deviation Dₘ = 40°, the refractive index is: Moderate
A) 1.33
B) 1.532
C) 1.44
D) 1.62
n = sin[(60+40)/2] / sin[60/2] = sin50° / sin30° = 0.766 / 0.5 = 1.532. Option B. (NCERT Ex 9.6)
Q17. Inside a prism at minimum deviation, the refracted ray is: Easy
A) Perpendicular to the base
B) Parallel to the base
C) At 45° to the base
D) At 30° to the base
At minimum deviation, i = e and r₁ = r₂. The refracted ray inside the prism becomes parallel to the base. Option B.
Q18. A thin prism of angle 4° and n = 1.5 causes a minimum deviation of: Moderate
A) 4°
B) 2°
C) 6°
D) 1°
For thin prism: Dₘ = (n−1)A = (1.5−1)×4° = 0.5×4° = 2°. Option B.
🔬
9.7.1 The Microscope
Simple microscope | Compound microscope | Magnifying power
Simple Microscope
🔍 Simple Magnifier
A converging lens of short focal length. Object placed within or at focal length. m = 1 + D/f (image at near point) m = D/f (image at infinity, relaxed eye) D = 25 cm (least distance of distinct vision)
Compound
🔬 Compound Microscope
Objective (short fₒ) + Eyepiece (short fₑ). Objective forms real, magnified image. Eyepiece acts as simple magnifier. m = mₒ × mₑ = (L/fₒ)(D/fₑ) L = tube length between focal points.
Key Point
👁️ Why Short Focal Lengths?
For large magnification both objective and eyepiece must have very short focal lengths. Tube length L must be large. In practice fₒ and fₑ ≈ 1–2 cm minimum due to practical limitations.
Simple Microscope
m = 1 + D/f (near point) | m = D/f (infinity)
Compound Microscope
m = mₒ × mₑ = (L/fₒ) × (D/fₑ) (image at infinity)
🖼️[NCERT Fig. 9.23: Simple microscope (a,b,c) | Fig. 9.24: Compound microscope ray diagram]
Add microscope diagrams here
🎯 Practice MCQs — Section 9.7.1
Q19. A simple microscope has f = 5 cm. Magnification when image forms at near point (D = 25 cm) is: Easy
A) 4
B) 5
C) 6
D) 25
m = 1 + D/f = 1 + 25/5 = 1 + 5 = 6. Option C.
Q20. Compound microscope: fₒ = 1 cm, fₑ = 2 cm, L = 20 cm, D = 25 cm. Magnification is: Hard
A) 100
B) 150
C) 250
D) 50
m = (L/fₒ)(D/fₑ) = (20/1)(25/2) = 20 × 12.5 = 250. Option C. (NCERT example)
Q21. In a compound microscope, the objective lens forms a: Easy
A) Virtual, erect, magnified image
B) Real, inverted, magnified image
C) Real, erect, diminished image
D) Virtual, inverted, diminished image
The objective lens forms a real, inverted, magnified image of the object. This image then serves as the object for the eyepiece. Option B.
🔭
9.7.2 Telescope
Refracting | Reflecting (Cassegrain) | Magnifying power
Refracting Telescope
🔭 Working Principle
Objective: large focal length fₒ, large aperture. Eyepiece: short focal length fₑ. m = fₒ/fₑ (normal adjustment) Tube length = fₒ + fₑ. Used for distant astronomical objects.
Reflecting Telescope
🪞 Reflecting / Cassegrain
Uses concave mirror instead of lens as objective. No chromatic aberration. Mirror weighs less, can be supported fully. India's largest: 2.34 m at Kavalur, Tamil Nadu.
Magnifying Power
📐 Telescope Formulas
m = fₒ/fₑ (image at infinity) m = (fₒ/fₑ)(1 + fₑ/D) (image at near point) To increase m: increase fₒ OR decrease fₑ. Large objective → more light gathering.
Telescope Magnifying Power
m = β/α = fₒ/fₑ (normal adjustment, image at infinity)
Tube Length
L = fₒ + fₑ
Resolving Power
Depends on aperture (diameter) of objective lens/mirror — larger aperture → better resolution.
Q22. A telescope has objective fₒ = 100 cm, eyepiece fₑ = 1 cm. Magnifying power (normal adjustment) is: Easy
A) 10
B) 50
C) 100
D) 1000
m = fₒ/fₑ = 100/1 = 100. Option C.
Q23. A telescope (fₒ = 144 cm, fₑ = 6 cm). Separation between objective and eyepiece in normal adjustment is: Moderate
A) 138 cm
B) 144 cm
C) 150 cm
D) 6 cm
Tube length = fₒ + fₑ = 144 + 6 = 150 cm. (NCERT Ex 9.13) Option C.
Q24. Which type of telescope has NO chromatic aberration? Easy
A) Refracting telescope
B) Galilean telescope
C) Reflecting telescope (concave mirror)
D) Simple telescope
Reflecting telescopes use concave mirrors. Mirrors reflect all wavelengths equally so there is no chromatic aberration. They are also lighter and easier to support. Option C.
📋
Chapter 9 — Key Summary
All important formulas at a glance for AAI ATC exam
🪞 Mirror Equation
1/v + 1/u = 1/f m = −v/u f = R/2 (−ve for concave)
💧 Snell's Law
n₁ sin i = n₂ sin r n = c/v Apparent depth = h/n
💎 TIR
sin iᶜ = n_rarer/n_denser Condition: denser→rarer, i > iᶜ Used in optical fibres