Top 10 Ray Optics Questions from Last 10 Years JEE (Main + Advanced)

Why These 10 Questions Are Crucial

Based on thorough analysis of JEE Main & Advanced papers (2014-2024), these 10 Ray Optics questions represent key patterns and concepts that frequently appear:

Mirror formula applications with sign convention twists
Lens combinations and equivalent focal length
Prism problems with minimum deviation concept
Optical instruments - microscope and telescope
Mixed concept problems combining ray optics with modern physics

JEE Advanced 2022 Medium

Problem 1: Concave Mirror with Liquid

A point object is placed at the center of curvature of a concave mirror of focal length 20 cm. A glass slab (μ = 1.5) of thickness 6 cm is inserted between the object and mirror. Find the position of final image.

Key Concept: Apparent depth + Mirror formula combination

Solution Approach:

Step 1: Object at center of curvature (u = -40 cm for mirror)

Step 2: Apparent shift due to slab = $t(1-\frac{1}{\mu}) = 6(1-\frac{1}{1.5}) = 2$ cm

Step 3: Effective object distance for mirror = 40 + 2 = 42 cm

Step 4: Apply mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Step 5: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-42}$ ⇒ $v = -52.5$ cm

Final Answer: 52.5 cm in front of mirror

JEE Advanced 2021 Hard

Problem 2: Lens-Mirror Combination

A convex lens (f = 15 cm) and a concave mirror (f = 20 cm) are placed co-axially 40 cm apart. An object is placed 30 cm in front of the lens. Find the position and nature of final image.

Key Concept: Multiple refraction/reflection with careful sign convention

Solution Approach:

Step 1: First refraction through lens:

$\frac{1}{15} = \frac{1}{v_1} - \frac{1}{-30}$ ⇒ $v_1 = 30$ cm (right of lens)

Step 2: This image acts as object for mirror:

Object distance for mirror = 40 - 30 = 10 cm

$\frac{1}{-20} = \frac{1}{v_2} + \frac{1}{-10}$ ⇒ $v_2 = -20$ cm

Step 3: This image acts as object for lens again:

Object distance = 40 - 20 = 20 cm

$\frac{1}{15} = \frac{1}{v_3} - \frac{1}{-20}$ ⇒ $v_3 = 8.57$ cm

Final Answer: 8.57 cm right of lens, real and inverted

JEE Main 2023 Medium

Problem 3: Prism with Grazing Incidence

A light ray is incident at grazing incidence on one face of a prism (A = 60°, μ = √2). Find the angle of emergence and deviation.

Key Concept: Grazing incidence → r₁ = critical angle, application of prism formula

Solution Approach:

Step 1: For grazing incidence, i₁ = 90°, so sin i₁ = 1

Step 2: Using Snell's law at first face: $1 \cdot \sin 90° = \mu \sin r_1$

$\sin r_1 = \frac{1}{\sqrt{2}}$ ⇒ $r_1 = 45°$

Step 3: $r_2 = A - r_1 = 60° - 45° = 15°$

Step 4: For second face: $\mu \sin r_2 = \sin i_2$

$\sqrt{2} \sin 15° = \sin i_2$ ⇒ $i_2 = \sin^{-1}(0.366) ≈ 21.5°$

Step 5: Deviation δ = i₁ + i₂ - A = 90° + 21.5° - 60° = 51.5°

Final Answer: Emergence angle = 21.5°, Deviation = 51.5°

🚀 Ray Optics Problem-Solving Strategies

Sign Convention Rules:

Object distance (u): always negative for real objects
Focal length: positive for convex lens/concave mirror
Image distance: positive for real, negative for virtual
Height: positive above principal axis

Common Pitfalls:

Forgetting apparent depth in slab problems
Missing multiple reflections in combinations
Confusing lens and mirror formulas
Not checking physical feasibility of answers

Problems 4-10 Available in Full Version

Includes 7 more crucial JEE Ray Optics problems with detailed solutions and concept maps

📚 Must-Know Ray Optics Formulas

Mirror Formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

with sign convention

Lens Formula:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Lens maker's formula

Prism Formula:

$$\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

at minimum deviation

Magnification:

$$m = -\frac{v}{u} = \frac{h_i}{h_o}$$

for both lenses and mirrors

📝 Quick Self-Test

Try these similar problems to test your Ray Optics understanding:

1. A convex lens (f=10cm) and concave mirror (f=15cm) are 25cm apart. Object at 20cm from lens. Find final image position.

2. Light passes through 60° prism with μ=1.5. If deviation is 40°, find angle of incidence.

3. Concave mirror (f=12cm) with water (μ=4/3) filled up to 9cm. Find image position of center of curvature object.

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