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Reflection Mirrors Refraction TIR Lenses Prisms Instruments Eye Defects
Chapter 21 Optics • Class 12

Ray Optics JEE Main & Advanced 2025-26

Master Reflection, Refraction, Mirrors, Lenses, Prisms & Optical Instruments with complete notes, formulas, ray diagrams, and 100+ solved problems. All JEE shortcuts included.

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Ray Optics JEE notes, Formulas, PYQs - Reflection, Refraction, Mirrors, Lenses
Ray Optics JEE Notes — Reflection, Refraction, Mirrors, Lenses, Prisms & Optical Instruments
1

Reflection of Light

Reflection is the phenomenon of bouncing back of light when it strikes a polished surface like a mirror. Understanding the laws of reflection is the foundation of Ray Optics and is essential for solving problems on plane mirrors and spherical mirrors.

1.1 Laws of Reflection

Two Laws of Reflection

Law 1:

The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence all lie in the same plane.

Law 2:

The angle of incidence (i) is always equal to the angle of reflection (r).

\[\angle i = \angle r\]

These laws are valid for all types of surfaces — plane, curved, smooth, or rough.

1.2 Types of Reflection

Regular (Specular) Reflection

When parallel light rays fall on a smooth surface, the reflected rays are also parallel.

  • Occurs on polished surfaces
  • Forms clear images
  • Example: Plane mirror, still water
Diffuse (Irregular) Reflection

When parallel light rays fall on a rough surface, the reflected rays scatter in different directions.

  • Occurs on rough surfaces
  • No clear image formed
  • Example: Wall, paper, ground

1.3 Plane Mirror Properties

Key Properties of Plane Mirror Image
  • ✓ Image is virtual and erect
  • ✓ Image size = Object size (m = 1)
  • ✓ Image distance = Object distance
  • ✓ Image is laterally inverted
  • ✓ Focal length f = ∞
  • ✓ Power P = 0
  • ✓ If mirror rotates by θ, reflected ray rotates by 2θ
  • ✓ Minimum mirror height = h/2 for full image

💡 JEE Important Formula

Number of images formed by two plane mirrors inclined at angle θ:

\[\text{If } \frac{360°}{\theta} = n\] \[\text{If n is even: Number of images} = n - 1\] \[\text{If n is odd: Number of images} = n - 1 \text{ (object not on bisector)}\] \[\text{If n is odd: Number of images} = n \text{ (object on bisector — but not applicable, still } n-1 \text{)}\]

When 360/θ is not integer, take the integer part for number of images.

📝 Solved Example 1

Question: Two plane mirrors are inclined at 60° to each other. An object is placed between them. Find the number of images formed.

Solution:

Given: θ = 60°

\[n = \frac{360°}{60°} = 6 \text{ (even)}\]

Since n is even:

\[\text{Number of images} = n - 1 = 6 - 1 = 5\]
2

Spherical Mirrors

Spherical mirrors are mirrors whose reflecting surface forms a part of a hollow sphere. They are of two types — concave (converging) and convex (diverging). Understanding mirror formula, magnification, and image formation is crucial for JEE.

2.1 Sign Convention (New Cartesian)

⚠️ CRITICAL: Master This First!

The sign convention is the #1 reason students get wrong answers. Memorize these rules:

  • 📍 All distances measured from Pole (P)
  • ➡️ Distances along incident light (right) → Positive (+)
  • ⬅️ Distances against incident light (left) → Negative (−)
  • ⬆️ Heights above principal axis → Positive (+)
  • ⬇️ Heights below principal axis → Negative (−)

Quick Reference:

  • • Object is always on left → u is always negative
  • • Concave mirror: f = negative, R = negative
  • • Convex mirror: f = positive, R = positive

2.2 Important Terms

Term Symbol Description
Pole P Center of reflecting surface (reference point)
Center of Curvature C Center of the sphere of which mirror is a part
Radius of Curvature R Distance from P to C (R = 2f)
Focus F Point where parallel rays converge/appear to diverge
Focal Length f Distance from P to F (f = R/2)
Principal Axis — Line joining P and C

2.3 Mirror Formula & Magnification

Mirror Formula
\[\boxed{\frac{1}{f} = \frac{1}{v} + \frac{1}{u}}\]

f = focal length

v = image distance from pole

u = object distance from pole

Magnification
\[\boxed{m = \frac{h'}{h} = -\frac{v}{u}}\]

m > 0 → Image is erect (virtual)

m < 0 → Image is inverted (real)

|m| > 1 → Magnified

|m| < 1 → Diminished

2.4 Image Formation by Concave Mirror

Object Position Image Position Image Nature Image Size
At ∞ At F Real, Inverted Highly diminished (point)
Beyond C Between F and C Real, Inverted Diminished
At C At C Real, Inverted Same size
Between F and C Beyond C Real, Inverted Magnified
At F At ∞ Real, Inverted Highly magnified
Between P and F Behind mirror Virtual, Erect Magnified

💡 JEE Memory Trick

Convex mirror ALWAYS forms: Virtual, Erect, Diminished image (behind the mirror)

Remember "VED" = Virtual, Erect, Diminished (for convex mirror, always!)

📝 Solved Example 2 (JEE Main Pattern)

Question: An object is placed 40 cm in front of a concave mirror of focal length 15 cm. Find the position, nature, and magnification of the image.

Solution:

Given: u = −40 cm, f = −15 cm (concave mirror)

Step 1: Apply mirror formula

\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
\[\frac{1}{-15} = \frac{1}{v} + \frac{1}{-40}\]
\[\frac{1}{v} = \frac{1}{-15} + \frac{1}{40} = \frac{-8 + 3}{120} = \frac{-5}{120}\]
\[v = -24 \text{ cm}\]

Step 2: Find magnification

\[m = -\frac{v}{u} = -\frac{-24}{-40} = -0.6\]
\[\text{Image: Real, Inverted, Diminished at 24 cm in front of mirror}\] \[v = -24\text{ cm}, \quad m = -0.6\]

2.5 Velocity of Image in Mirrors

Image Velocity Formula (Along Principal Axis)

Differentiating mirror formula with respect to time:

\[\boxed{v_{\text{image}} = -\frac{v^2}{u^2} \cdot v_{\text{object}}}\]

This is the velocity of image relative to mirror along the principal axis.

For Perpendicular to Axis:

\[v_{\text{image}}^{\perp} = -\frac{v}{u} \cdot v_{\text{object}}^{\perp}\]
3

Refraction of Light

Refraction is the bending of light when it passes from one transparent medium to another. This occurs because the speed of light changes in different media. Snell's Law is the fundamental equation governing refraction.

3.1 Snell's Law of Refraction

\[\boxed{n_1 \sin\theta_1 = n_2 \sin\theta_2}\]
n₁ = RI of medium 1
n₂ = RI of medium 2
θ₁ = Angle of incidence
θ₂ = Angle of refraction

Refractive Index:

\[n = \frac{c}{v} = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in medium}}\]

Refractive index of vacuum/air = 1, Water ≈ 1.33, Glass ≈ 1.5, Diamond ≈ 2.42

3.2 Key Concepts in Refraction

Apparent Depth
\[\text{Apparent Depth} = \frac{\text{Real Depth}}{n}\]

Object in denser medium, observer in rarer medium

Shift: d = t(1 - 1/n)

where t = thickness of slab

Lateral Shift (Glass Slab)
\[d = \frac{t \sin(i - r)}{\cos r}\]

where t = thickness, i = angle of incidence, r = angle of refraction

Note: Emergent ray is parallel to incident ray but laterally displaced

3.3 Refraction at Spherical Surfaces

Refraction Formula at Single Spherical Surface
\[\boxed{\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}}\]

n₁ = RI of medium where object is placed

n₂ = RI of medium where image is formed

R = Radius of curvature of spherical surface

📝 Solved Example 3

Question: A coin is placed at the bottom of a tank filled with water (n = 4/3) to a depth of 80 cm. Find the apparent position of the coin when viewed from directly above.

Solution:

Given: Real depth = 80 cm, n = 4/3

\[\text{Apparent Depth} = \frac{\text{Real Depth}}{n} = \frac{80}{4/3} = 80 \times \frac{3}{4} = 60 \text{ cm}\]
\[\text{Apparent depth} = 60 \text{ cm (coin appears 20 cm closer)}\]
4

Total Internal Reflection (TIR)

Total Internal Reflection is the complete reflection of light back into the denser medium when it strikes the boundary at an angle greater than the critical angle. This is one of the most important and frequently asked topics in JEE from Ray Optics.

4.1 Conditions for TIR

Two Necessary Conditions:

Condition 1:

Light must travel from denser medium to rarer medium (n₁ > n₂)

Condition 2:

Angle of incidence must be greater than critical angle (i > C)

4.2 Critical Angle

The critical angle C is the angle of incidence for which angle of refraction = 90°

\[\boxed{\sin C = \frac{n_2}{n_1} = \frac{1}{n} \text{ (when rarer medium is air)}}\]

Critical Angles of Common Materials:

Water

C ≈ 49°

Glass

C ≈ 42°

Diamond

C ≈ 24.4°

Dense Flint

C ≈ 33°

4.3 Applications of TIR

Optical Fiber
  • Uses principle of TIR
  • Light enters core and undergoes repeated TIR
  • Core has higher RI than cladding
  • Used in telecommunications, endoscopy
Other Applications
  • Mirages: Due to TIR in hot air layers
  • Brilliance of diamonds: Small critical angle
  • Totally reflecting prisms: 45-90-45° prisms
  • Sparkling of water fountain

📝 Solved Example 4 (JEE Advanced Pattern)

Question: A point source of light is placed at the bottom of a water tank (n = 4/3). Find the radius of the bright circle seen from above the water surface if the depth of water is 80 cm.

Solution:

Given: h = 80 cm, n = 4/3

Step 1: Find critical angle

\[\sin C = \frac{1}{n} = \frac{3}{4} \implies \tan C = \frac{3}{\sqrt{7}}\]

Step 2: Radius of bright circle

\[R = h \tan C = 80 \times \frac{3}{\sqrt{7}}\]
\[R = \frac{240}{\sqrt{7}} \approx 90.7 \text{ cm}\]

General formula: R = h/√(n² - 1)

5

Thin Lenses

A lens is a transparent optical medium bounded by two surfaces, at least one of which is curved. Lenses are the most important topic in Ray Optics for JEE — expect at least 1 question from this section every year.

5.1 Lens Formula

Thin Lens Formula
\[\boxed{\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}\]

⚠️ Note: Different from mirror formula!

Mirror: 1/f = 1/v + 1/u

Lens: 1/f = 1/v − 1/u

Magnification
\[\boxed{m = \frac{h'}{h} = \frac{v}{u}}\]

⚠️ No negative sign for lens!

Mirror: m = −v/u

Lens: m = v/u

5.2 Lens Maker's Formula

\[\boxed{\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}\]
μ = Refractive index of lens
R₁ = Radius of 1st surface
R₂ = Radius of 2nd surface
f = Focal length

When lens is in a medium of RI μₘ:

\[\frac{1}{f} = \left(\frac{\mu_L}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]

5.3 Power of Lens

\[\boxed{P = \frac{1}{f \text{ (in meters)}} \quad \text{Unit: Dioptre (D)}}\]

Convex lens:

f > 0 → P > 0 (converging)

Concave lens:

f < 0 → P < 0 (diverging)

5.4 Combination of Lenses

Lenses in Contact
\[\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \cdots\]
\[P_{eq} = P_1 + P_2 + P_3 + \cdots\]
\[m_{total} = m_1 \times m_2 \times m_3 \times \cdots\]

📝 Solved Example 5 (JEE Main 2023 Type)

Question: Two thin lenses of focal lengths +20 cm and −10 cm are placed in contact. Find the power and focal length of the combination. What type of combination is it?

Solution:

Given: f₁ = +20 cm = 0.2 m, f₂ = −10 cm = −0.1 m

Step 1: Find individual powers

\[P_1 = \frac{1}{0.2} = +5\text{ D}, \quad P_2 = \frac{1}{-0.1} = -10\text{ D}\]

Step 2: Combined power

\[P = P_1 + P_2 = 5 + (-10) = -5\text{ D}\]

Step 3: Focal length

\[f = \frac{1}{P} = \frac{1}{-5} = -0.2\text{ m} = -20\text{ cm}\]
\[\text{P = −5 D, f = −20 cm (Diverging combination)}\]

5.5 Image Formation by Convex Lens

Object Position Image Position Image Nature Image Size
At ∞ At F₂ Real, Inverted Point sized
Beyond 2F₁ Between F₂ and 2F₂ Real, Inverted Diminished
At 2F₁ At 2F₂ Real, Inverted Same size
Between F₁ and 2F₁ Beyond 2F₂ Real, Inverted Magnified
At F₁ At ∞ Real, Inverted Highly magnified
Between O and F₁ Same side as object Virtual, Erect Magnified

💡 Concave Lens — Always Remember

Concave lens ALWAYS forms: Virtual, Erect, Diminished image (on same side as object)

Similar to convex mirror — both always give VED images!

6

Prisms & Dispersion of Light

A prism is a transparent optical element with flat, polished surfaces that refract light. The most common type is a triangular prism. Prisms are used to disperse white light into its component colors and to deviate light rays.

6.1 Prism Formulas

Angle of Prism:

\[\boxed{A = r_1 + r_2}\]

Angle of Deviation:

\[\boxed{\delta = i_1 + i_2 - A}\]

At Minimum Deviation (δₘ):

\[i_1 = i_2 = i, \quad r_1 = r_2 = r = A/2\]
\[\boxed{\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}}\]

This is the most important prism formula for JEE!

For Thin Prism (A < 10°):

\[\boxed{\delta = (\mu - 1)A}\]

Very useful for quick calculations in JEE!

6.2 Dispersion of Light

What is Dispersion?

Splitting of white light into its component colors by a prism due to different refractive indices for different wavelengths.

VIBGYOR Order (increasing λ):

Violet → Indigo → Blue → Green → Yellow → Orange → Red

μ_violet > μ_red → δ_violet > δ_red

Angular Dispersion & Dispersive Power

Angular Dispersion:

\[\theta = \delta_v - \delta_r = (\mu_v - \mu_r)A\]

Dispersive Power:

\[\omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{\mu_v - \mu_r}{\mu_y - 1}\]

⚠️ JEE Advanced Concept: Deviation Without Dispersion

Two prisms can be combined to produce deviation without dispersion (achromatic combination):

\[\omega_1 \delta_1 + \omega_2 \delta_2 = 0\]

Or for dispersion without deviation:

\[\delta_1 + \delta_2 = 0 \implies (\mu_1 - 1)A_1 = (\mu_2 - 1)A_2\]
7

Optical Instruments

Optical instruments use mirrors and lenses to create images of objects for various purposes — magnification, projection, etc. Understanding magnifying power formulas is key for JEE.

7.1 Magnifying Power Formulas

Instrument At D (25 cm) At ∞ (Normal Adjustment)
Simple Microscope m = 1 + D/f m = D/f
Compound Microscope m = (v₀/u₀)(1 + D/fₑ) m = (v₀/u₀)(D/fₑ) ≈ (L·D)/(f₀·fₑ)
Astronomical Telescope m = (f₀/fₑ)(1 + fₑ/D) m = f₀/fₑ
Galilean Telescope — m = f₀/fₑ

💡 Key Points for JEE

  • D = 25 cm (least distance of distinct vision for normal eye)
  • Tube length of telescope (normal adjustment): L = f₀ + fₑ
  • Tube length of compound microscope: L = v₀ + |uₑ|
  • For higher magnification in telescope → large f₀, small fₑ
  • Astronomical telescope gives inverted image; Galilean gives erect image

📝 Solved Example 6

Question: An astronomical telescope in normal adjustment has a magnifying power of 20. If the focal length of the objective is 100 cm, find the focal length of the eyepiece and the length of the telescope.

Solution:

Given: m = 20, f₀ = 100 cm (normal adjustment)

Step 1: Find fₑ

\[m = \frac{f_0}{f_e} \implies f_e = \frac{f_0}{m} = \frac{100}{20} = 5 \text{ cm}\]

Step 2: Length of telescope

\[L = f_0 + f_e = 100 + 5 = 105 \text{ cm}\]
\[f_e = 5 \text{ cm}, \quad L = 105 \text{ cm}\]
8

Defects of Vision & Human Eye

The human eye is a remarkable optical instrument. It uses a convex lens (crystalline lens) to form real images on the retina. Understanding eye defects and their corrections is important for JEE.

8.1 The Human Eye

  • ✓ Near point (D): 25 cm (closest distance for clear vision)
  • ✓ Far point: ∞ (farthest distance for clear vision)
  • ✓ Accommodation: Ability to adjust focal length by changing lens curvature
  • ✓ Range of vision: 25 cm to ∞ (for normal eye)
  • ✓ Persistence of vision: 1/16 second

8.2 Common Eye Defects

Defect Problem Correction
Myopia (Short-sightedness) Cannot see distant objects; image forms before retina; far point < ∞ Concave lens (diverging); f = −far point
Hypermetropia (Long-sightedness) Cannot see nearby objects; image forms behind retina; near point > 25 cm Convex lens (converging)
Presbyopia Weakening of ciliary muscles with age; both near and far vision affected Bifocal lens (concave + convex)
Astigmatism Cornea not perfectly spherical; different focal lengths in different planes Cylindrical lens

📝 Solved Example 7

Question: A person can see objects clearly only up to 200 cm. What lens should be used to correct the vision, and what should be its power?

Solution:

This is Myopia (far point = 200 cm instead of ∞)

Need concave lens with f = −far point = −200 cm = −2 m

\[P = \frac{1}{f} = \frac{1}{-2} = -0.5 \text{ D}\]
\[\text{Concave lens with Power = −0.5 D}\]

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

  • ✓ Mirror & Lens Problems: 30%
  • ✓ Total Internal Reflection: 20%
  • ✓ Prism & Dispersion: 20%
  • ✓ Optical Instruments: 15%
  • ✓ Refraction at Surfaces: 15%

JEE Advanced (Last 5 Years)

  • ✓ Combination of lenses/mirrors: 25%
  • ✓ Refraction at curved surfaces: 25%
  • ✓ TIR based problems: 20%
  • ✓ Prism with multiple concepts: 15%
  • ✓ Image velocity problems: 15%

Top 10 Most Repeated Question Types

  1. Mirror formula & magnification numerical problems
  2. Lens formula with sign convention applications
  3. Total Internal Reflection — critical angle calculations
  4. Combination of thin lenses (equivalent focal length, power)
  5. Prism minimum deviation problems
  6. Refraction at spherical surfaces (single surface formula)
  7. Apparent depth/shift problems
  8. Magnifying power of telescope and microscope
  9. Eye defects and corrective lens power
  10. Image velocity in mirrors (along/perpendicular to axis)

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)
JEE Advanced: 6-12 marks (2-3 questions)
Difficulty Level: Medium to Hard
Time Required: 4-5 hours for mastery
Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

  1. An object is placed 30 cm from a concave mirror of focal length 20 cm. Find the image position and magnification.
  2. Two plane mirrors are placed at 72° to each other. Find the number of images formed.
  3. A convex lens of focal length 15 cm forms an image at 30 cm. Find the object distance.
  4. Find the critical angle for water-air interface (n_water = 4/3).
  5. A coin is placed 12 cm below the surface of water (n = 4/3). Find the apparent depth.
  6. Two lenses of power +5 D and −3 D are placed in contact. Find the equivalent focal length.
  7. Find the minimum height of a plane mirror required to see full image of a person of height 180 cm.
  8. A convex mirror has a focal length of 20 cm. An object is placed at 30 cm. Find image position.

Level 2: Intermediate (JEE Main/Advanced)

  1. A glass slab of thickness 10 cm and refractive index 1.5 is placed on a table. Find the apparent shift of a mark on the table.
  2. A prism of angle 60° has a refractive index of √2. Find the angle of minimum deviation.
  3. An equiconvex lens of focal length 20 cm is cut in half. Find the focal length of each half.
  4. A concave mirror and convex lens both have focal length 15 cm. Object is at 20 cm. Compare images formed.
  5. Light enters an optical fiber (n_core = 1.5, n_cladding = 1.2). Find the maximum angle of acceptance.
  6. A compound microscope has f₀ = 1 cm, fₑ = 5 cm, tube length = 20 cm. Find magnifying power at D.
  7. Two thin prisms of angles 4° and 6° (μ = 1.5 each) are combined. Find net deviation and dispersion.
  8. Object moving at 5 cm/s towards a concave mirror (f = 10 cm) when at 15 cm. Find image velocity.

Level 3: Advanced (JEE Advanced/Olympiad)

  1. A convex lens of focal length 20 cm is placed in contact with a concave mirror of focal length 15 cm. An object is at 30 cm from this system. Find the final image.
  2. A point source is placed at the bottom of a tank containing water (n = 4/3) of depth h. Find the radius of the bright patch visible from above.
  3. A fish in a lake sees a bird flying at height H above the surface. If the actual height is H and fish is at depth d, find the apparent height of the bird as seen by fish.
  4. A lens of refractive index 1.5 in air has power P. It is immersed in water (n = 4/3). Find new power in terms of P.
  5. A silvered lens (planoconvex lens with silvered plane surface) has R = 30 cm and μ = 1.5. Find equivalent mirror focal length.
  6. Two thin prisms are combined to produce dispersion without deviation. If first prism has A = 4°, μ_v = 1.56, μ_r = 1.52, and second prism has μ_v = 1.68, μ_r = 1.64, find angle of second prism.
  7. A lens has different radii R₁ = 20 cm, R₂ = 30 cm and μ = 1.5. It is placed in a medium of μ = 1.6. Find its nature (converging/diverging) and focal length.
  8. An object is placed 20 cm in front of a glass slab of thickness 12 cm (μ = 1.5), behind which a concave mirror of focal length 10 cm is placed. Find the position of the final image.
Get Solutions PDF with Step-by-Step Working

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Ray Optics - Complete Guide for JEE 2025-26

Why Ray Optics is Important for JEE?

Ray Optics is one of the highest-scoring chapters in JEE Physics. It carries 5-6% weightage with formula-based questions that are easy to solve if concepts are clear:

  • Mirror & Lens formulas - Direct numerical problems (most common)
  • Total Internal Reflection - Concept-based MCQs
  • Prism Problems - Minimum deviation calculations
  • Optical Instruments - Magnifying power formulas

In JEE Advanced, ray optics questions often involve multi-concept problems combining mirrors, lenses, and refraction at surfaces.

Key Formulas to Remember

1. Must-Know Formulas

  • • Mirror: 1/f = 1/v + 1/u, m = −v/u
  • • Lens: 1/f = 1/v − 1/u, m = v/u
  • • Snell's Law: n₁sinθ₁ = n₂sinθ₂
  • • Lens Maker's: 1/f = (μ−1)(1/R₁ − 1/R₂)
  • • Power: P = 1/f (in meters)
  • • Prism: μ = sin[(A+δₘ)/2] / sin(A/2)

2. Quick Shortcuts

  • • Thin prism: δ = (μ−1)A
  • • Apparent depth = Real depth / n
  • • Critical angle: sinC = 1/n
  • • Telescope (normal): m = f₀/fₑ

📚 How to Study Ray Optics Effectively?

For JEE Main Students:

  1. Time Required: 4-5 days (2-3 hours/day)
  2. Master sign convention first — it's the foundation
  3. Practice 50+ numericals on mirror & lens formula
  4. Memorize image formation tables for mirrors & lenses
  5. Solve 20+ TIR and prism problems
  6. Learn magnifying power formulas for all instruments

For JEE Advanced Students:

  1. Time Required: 7-8 days (3 hours/day)
  2. Master refraction at curved surfaces formula
  3. Practice silvered lens and combination problems
  4. Solve multi-concept problems (mirror + lens + slab)
  5. Study image velocity in mirrors and lenses
  6. Practice deviation without dispersion problems

⚠️ Common Mistakes to Avoid in JEE Exam

  • ❌
    Wrong sign convention: Object distance u is ALWAYS negative (object on left). Most students forget this and get wrong answers.
  • ❌
    Confusing mirror and lens formulas: Mirror: 1/f = 1/v + 1/u (plus sign). Lens: 1/f = 1/v − 1/u (minus sign). Similarly, m = −v/u (mirror) vs m = v/u (lens).
  • ❌
    Unit conversion in Power: P = 1/f requires f in METERS, not centimeters. f = 20 cm → P = 1/0.2 = +5D, NOT 1/20.
  • ❌
    TIR conditions: TIR only happens from denser to rarer medium. Many students try to apply it from rarer to denser — not possible!
  • ❌
    Ignoring lateral inversion: Convex mirror always gives virtual erect diminished image. Don't confuse with concave mirror properties.

📊 JEE Previous Year Question Analysis (2015-2024)

Year JEE Main JEE Advanced Topic Focus
2024 2 Questions (8 marks) 2 Questions (8 marks) Lens combination, TIR, Prism deviation
2023 1 Question (4 marks) 3 Questions (10 marks) Mirror formula, Refraction at curved surface
2022 2 Questions (8 marks) 2 Questions (7 marks) Silvered lens, Optical fiber, Telescope

Trend: JEE Advanced increasingly asks multi-step problems combining reflection, refraction, and lens concepts. Pure formula-based questions are becoming rare in Advanced — expect application-based problems.