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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

If the critical angle for total internal reflection from a medium to vacuum is 4545^{\circ}, the velocity of light in the medium is:

A

1.5×108 m/s1.5 \times 10^8 \text{ m/s}

B

32×108 m/s\frac{3}{\sqrt{2}} \times 10^8 \text{ m/s}

C

2×108 m/s\sqrt{2} \times 10^8 \text{ m/s}

D

3×108 m/s3 \times 10^8 \text{ m/s}

Step-by-Step Solution

  1. Critical Angle Relation: The refractive index (μ\mu) of a medium is related to the critical angle (CC) for total internal reflection by the formula: μ=1sinC\mu = \frac{1}{\sin C}.
  2. Calculate Refractive Index: Given C=45C = 45^{\circ}. μ=1sin45=11/2=2\mu = \frac{1}{\sin 45^{\circ}} = \frac{1}{1/\sqrt{2}} = \sqrt{2}
  3. Velocity Relation: The velocity of light in a medium (vv) is related to the speed of light in vacuum (cc) and the refractive index (μ\mu) by: v=cμv = \frac{c}{\mu}.
  4. Calculation: Substituting c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and μ=2\mu = \sqrt{2}: v=3×1082 m/sv = \frac{3 \times 10^8}{\sqrt{2}} \text{ m/s}
  5. Conclusion: The velocity of light in the medium is 32×108 m/s\frac{3}{\sqrt{2}} \times 10^8 \text{ m/s}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTScriticalinternalreflectionmediumvacuum

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