Q.22 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 22 Solution

Topic: Ray Optics & Optical Instruments · Official NTA Booklet Code 50

Question

Consider three media P, Q and R with refractive indices 1, 1.25 and 1.5, respectively. The medium Q having a thickness of 5 cm is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is h₁. For similar observation from medium R, the apparent depth is h₂. The value of |h₁ − h₂|, in cm, is:

✓

Verified Against NTA Final Answer Key

Option (2) : |h_1 - h_2| = 1 cm

📚 Ray Optics & Optical Instruments Medium

📋 Given Parameters & Problem Setup

Medium P n_P = 1 (top medium)

Medium Q n_Q = 1.25 = 5/4, thickness t = 5 cm

Medium R n_R = 1.5 = 3/2 (bottom medium)

Object Position At the exact center of medium Q (real depth d = 2.5 cm from both top and bottom interfaces)

Apparent Depths h_1 viewed from P; h_2 viewed from R

Target Difference |h_1 - h_2| in cm

⚡

The 30-Second Exam Hall Shortcut: ⚡ Apparent Depth Ratio Method

⏱️ Prevents confusing real depth with total thickness.

Apparent depth = Real depth * (n_observer / n_object)

1 Real distance from center of Q to both surfaces is d = 5 cm / 2 = 2.5 cm.

2 Viewed from medium P (n_P = 1): h_1 = 2.5 * (n_P / n_Q) = 2.5 * (1 / 1.25) = 2.5 * (4/5) = 2.0 cm.

3 Viewed from medium R (n_R = 1.5): h_2 = 2.5 * (n_R / n_Q) = 2.5 * (1.5 / 1.25) = 2.5 * (6/5) = 3.0 cm.

4 Magnitude of difference: |h_1 - h_2| = |2.0 - 3.0| = 1.0 cm. Solved in 20 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Snell's Law for paraxial rays: n_1 * tan(theta_1) = n_2 * tan(theta_2) ==> h_apparent / d_real = n_observer / n_object.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determine the Real Depth of the Object from Both Interfaces

The slab of medium Q has a thickness of 5 cm. Since object O is located at its center:

d_1 = d_2 = \frac{t}{2} = \frac{5\text{ cm}}{2} = 2.5\text{ cm}

• d_1 is the real depth from the upper P-Q interface.
• d_2 is the real depth from the lower Q-R interface.

Step 2

Calculate Apparent Depth h_1 When Viewed from Medium P

Observer is in medium P (n_P = 1), looking at object O in medium Q (n_Q = 1.25):

h_1 = d_1 \left( \frac{n_P}{n_Q} \right) = 2.5 \left( \frac{1}{1.25} \right) = 2.5 \left( \frac{4}{5} \right) = 2.0\text{ cm}

• The object appears shifted closer to the interface by 0.5 cm.

Step 3

Calculate Apparent Depth h_2 When Viewed from Medium R

Observer is in medium R (n_R = 1.5), looking at object O in medium Q (n_Q = 1.25):

h_2 = d_2 \left( \frac{n_R}{n_Q} \right) = 2.5 \left( \frac{1.5}{1.25} \right) = 2.5 \left( \frac{6}{5} \right) = 3.0\text{ cm}

• Because n_R > n_Q, the object appears further from the interface.

Step 4

Compute the Absolute Difference |h_1 - h_2|

Subtract the two apparent depths:

|h_1 - h_2| = |2.0\text{ cm} - 3.0\text{ cm}| = 1.0\text{ cm}

• Matches Option (2) exactly.

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 Option (1)

✗ Incorrect

Verdict & Scientific Reason:

Symmetry fallacy.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes because the object is at the center of Q, it must appear at the same distance from both sides.

Note: Refractive indices of surrounding media are different (1 vs 1.5), producing asymmetric apparent depths.

2 Option (2)

✓ 100% Correct

Verdict & Scientific Reason:

100% mathematically verified; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Correctly applies paraxial refraction formula from both media: |2.0 - 3.0| = 1 cm.

Note: Exact optics.

3 Option (3)

✗ Incorrect

Verdict & Scientific Reason:

Used full thickness 5 cm.

⚠️

Student Fallacy (Why candidates pick this trap): Uses d = 5 cm instead of d = 2.5 cm: |4.0 - 6.0| = 2 cm.

Note: The object is placed at the CENTER of medium Q, so real depth is 2.5 cm, not 5 cm.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Took h_2 value alone.

⚠️

Student Fallacy (Why candidates pick this trap): Marks the value of h_2 = 3 cm instead of the difference |h_1 - h_2|.

Note: Always re-read what is requested: the DIFFERENCE |h_1 - h_2|.

⚠️

NTA Trap Alert: The Full Slab Thickness Oversight (Using 5 cm instead of 2.5 cm)

Many students use real depth d = 5 cm, calculating h_1 = 4 cm and h_2 = 6 cm, which leads to |h_1 - h_2| = 2 cm (Option 3).

🛡️ How to avoid the -1 mark: Underline the phrase "at the center of medium Q" — center means real depth is d = 5 / 2 = 2.5 cm!

📖

Core NEET Syllabus Standard

NEET UG Physics Core Standards · Ray Optics and Optical Instruments Standard Benchmark

Core Scientific Principle Tested:

"Apparent depth of an object observed from medium 1 inside medium 2: h_app = d_real * (n_1 / n_2)."

Syllabus Focus: Refraction at Plane Surfaces and Apparent Depth

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 22 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Apparent Depth Ratio Method

Step-by-Step Rigorous Workflow & Mathematical Proof

Determine the Real Depth of the Object from Both Interfaces

Calculate Apparent Depth h_1 When Viewed from Medium P

Calculate Apparent Depth h_2 When Viewed from Medium R

Compute the Absolute Difference |h_1 - h_2|

360° Option Elimination Engine Key Feature

NTA Trap Alert: The Full Slab Thickness Oversight (Using 5 cm instead of 2.5 cm)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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