Problem Types & Solved Examples
Every problem pattern you'll face in exams - decoded
The Pattern Recognition Method: 90% of Ray Optics problems fall into 6 categories. Master these patterns and you've mastered the chapter.
Type 1: Direct Formula Application
Pattern: Given 2 variables, find third using formula. 40% of NEET, 30% of JEE Main.
Problem: An object is placed 30 cm in front of a concave mirror of focal length 20 cm. Find the position and nature of image.
Given:
u = -30 cm (object always negative)
f = -20 cm (concave mirror, converging)
Find: v, nature
Step 1: Identify the concept
Mirror formula: 1/f = 1/v + 1/u
Step 2: Substitute values
1/v = -1/20 + 1/30
1/v = (-3 + 2)/60 = -1/60
v = -60 cm
Step 3: Interpret
v = -60 cm → Negative means real image (on same side as object, in front of mirror)
Object beyond C → Image between F and C ✓
Magnification:
Negative m → Inverted
|m| = 2 → Magnified 2x
Shortcut: Object beyond C in concave mirror → Real, inverted, diminished/magnified depending on exact position. Since 30 cm is between f(20) and R(40), image is magnified.
Problem: A convex lens of focal length 15 cm forms an image at 60 cm. Find object distance.
Given: f = +15 cm, v = +60 cm (real image on right)
Find: u
1/15 = 1/60 - 1/u
1/u = 1/60 - 1/15 = (1-4)/60 = -3/60 = -1/20
u = -20 cm
Students often write "u = 20 cm". Wrong! u is ALWAYS negative in standard setup.
Type 2: Conceptual Questions
Pattern: "Why", "Explain", "What happens if". Tests understanding. 20% of CBSE, 30% of JEE Advanced.
Problem: Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Concept Selection:
Properties of convex mirror: Always virtual, erect, diminished, wider field of view
Answer:
- Wider field of view: Convex mirror diverges light, allowing driver to see larger area
- Always forms image: Regardless of object distance, image is always formed between P and F
- Erect image: Easier to interpret what's behind
- Diminished: Fits more area in small mirror
CBSE loves this question. Give at least 3 reasons with proper physics terminology.
Problem: A lens behaves as a converging lens in air and diverging lens in water. What is the refractive index of lens material?
Concept Selection:
Lens maker's formula: 1/f = (n_lens/n_medium - 1)(1/R₁ - 1/R₂)
Nature changes when (n_lens/n_medium - 1) changes sign
Solution:
In air: n_medium = 1, lens is converging → n_lens > 1
In water: n_medium = 4/3, lens is diverging → n_lens < 4/3
Therefore: 1 < n_lens < 4/3
Possible range: 1 < n < 1.33
JEE Advanced loves such problems. Key insight: Behavior depends on relative refractive index, not absolute.
Type 3: Multi-Step Problems
Pattern: Lens combinations, compound systems. Output of one becomes input of next. 30% of JEE Main/Advanced.
Problem: Two thin lenses of focal lengths +10 cm and -5 cm are placed in contact. An object is placed 20 cm from the combination. Find image position.
Method 1: Combined Power (Faster)
P = 1/0.1 + 1/(-0.05) = 10 - 20 = -10 D
f_eq = -0.1 m = -10 cm
Now use lens formula:
1/(-10) = 1/v - 1/(-20)
1/v = -1/10 - 1/20 = -3/20
v = -6.67 cm
Nature: Virtual, erect, on same side as object (diverging combination)
Method 2: Step by Step (Conceptual)
First lens (f₁ = 10 cm):
This image acts as object for second lens at distance u₂ = -20 cm
For lenses in contact, Method 1 is faster. For separated lenses, use Method 2.
Type 4: Graph-Based Questions
Pattern: Plot v vs u, 1/v vs 1/u, etc. Analytical questions. 15% of JEE Main/Advanced.
Problem: For a convex lens, draw graph of image position (v) vs object position (u).
Key Points to Plot:
- u → -∞: v → f (image at focus)
- u = -2f: v = +2f (symmetry point)
- u = -f: v → ∞ (object at focus)
- u → 0⁻: v → -∞ (object close to lens)
Graph characteristics:
- Rectangular hyperbola shape
- Asymptotes at u = -f and v = f
- Two branches: v > 0 (real) and v < 0 (virtual)
Why hyperbola? Because 1/v = 1/f + 1/u is a linear relationship. When you invert, you get hyperbola.
Problem: Plot graph of magnification (m) vs object distance (u) for concave mirror.
From formula: m = -v/u and v = uf/(u+f)
Therefore: m = -f/(u+f)
Key observations:
- At u = -f: m → -∞ (infinite magnification)
- At u = -2f: m = -1 (same size, inverted)
- As u → -∞: m → 0 (highly diminished)
- For u between 0 and -f: m > 0 (erect, virtual)
Graph has vertical asymptote at u = -f
Type 5: Assertion & Reason
Pattern: NEET specialty. Two statements, find relationship. Master the 4 answer types.
Answer Code:
(A) Both true, R is correct explanation of A
(B) Both true, R is NOT correct explanation of A
(C) A is true, R is false
(D) A is false, R is true
Assertion (A): A convex mirror always produces a virtual image.
Reason (R): Convex mirror has positive focal length.
Analysis:
Assertion: TRUE. Convex mirror always produces virtual, erect, diminished images.
Reason: TRUE. Convex mirror has f > 0.
Is R correct explanation of A? YES. Because f > 0 and mirror formula 1/f = 1/v + 1/u with u < 0 always gives v > 0 (virtual).
Answer: (A)
Always check: (1) Is A true? (2) Is R true? (3) Does R explain A? Follow this order to avoid confusion.
Assertion (A): Total internal reflection can occur when light travels from glass to air.
Reason (R): Speed of light is higher in air than in glass.
Assertion: TRUE. Glass (denser) to air (rarer) can show TIR.
Reason: TRUE. Air has lower RI, so light travels faster.
Is R correct explanation? Partially. R states a fact but doesn't directly explain why TIR happens (needs critical angle concept).
Answer: (B) - Both true but R not proper explanation.
Type 6: Case-Based Questions
Pattern: CBSE 2023+ format. Paragraph + 4-5 MCQs. Tests application and reading comprehension.
Case Study: Telescope Design
An astronomical telescope is designed to observe distant celestial objects. The telescope has an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. The telescope is in normal adjustment (final image at infinity).
Q1: What is the magnifying power of the telescope?
(a) 5 (b) 10 (c) 20 (d) 25
Answer: (c) 20
In normal adjustment: m = -f₀/fₑ = -100/5 = -20 (magnitude = 20)
Q2: What is the length of telescope tube?
(a) 95 cm (b) 100 cm (c) 105 cm (d) 110 cm
Answer: (c) 105 cm
L = f₀ + fₑ = 100 + 5 = 105 cm
Q3: If we want to double the magnifying power, we should:
(a) Double f₀ (b) Halve fₑ (c) Double both (d) Halve both
Answer: (b) Halve fₑ
Since m = f₀/fₑ, to double m, we need to halve fₑ (easier than making objective very large)