Three media P, Q and R have refractive indices 1, 1.25 and 1.5 respectively. Medium Q of thickness 5 cm is placed between extended media P and R. An object O is at the centre of Q. Viewed from P (near normal) the apparent depth of O is h₁; viewed from R the apparent depth is h₂. The value of |h₁-h₂| (in cm) is:
- A.0
- B.1✓
- C.2
- D.3
Correct Answer
(B) 1
Solution & Explanation
The object O sits at the centre of slab Q (thickness 5 cm), so its real depth from EACH face is 5/2 = 2.5 cm. Apparent depth seen across a flat interface = real depth × (n of viewing medium)/(n of object medium). Viewed from P (n_P = 1) through Q (n_Q = 1.25): h₁ = 2.5 × (n_P/n_Q) = 2.5 × (1/1.25) = 2.0 cm. Viewed from R (n_R = 1.5) through Q: h₂ = 2.5 × (n_R/n_Q) = 2.5 × (1.5/1.25) = 2.5 × 1.2 = 3.0 cm. Difference: |h₁ − h₂| = |2.0 − 3.0| = 1 cm. Note: looking into a denser medium (R) makes the object appear deeper (h₂ > 2.5), into a rarer one (P) shallower — so the two depths differ by 1 cm, option B.
✅ We had this exact concept in our bank before the exam — Strong match, 90% match.
That's the Predicted Batch — it trains you on the exact high-probability concepts NEET repeats every year. 98% of the 2026 paper (176/180) was in our bank before the exam.
Master these concepts + 50,000+ NEET PYQs in the app — free, built for NEET 2027.
Start the Predicted Batch — Free →