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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

A horizontal ray of light is incident on a right-angled prism with a prism angle of 6∘6^{\circ}. If the refractive index of the material of the prism is 1.5, then the angle of emergence will be:

A

9∘9^{\circ}

B

10∘10^{\circ}

C

4∘4^{\circ}

D

6∘6^{\circ}

Step-by-Step Solution

  1. Ray Entry: A horizontal ray of light incident on the vertical face of a right-angled prism falls normally on it (angle of incidence i=0∘i = 0^{\circ}). It passes through undeviated, meaning the angle of refraction at the first face is r1=0∘r_1 = 0^{\circ}.
  2. Angle of Incidence at Second Face: For a prism, the angle of the prism AA is related to the internal angles of refraction by A=r1+r2A = r_1 + r_2. Given A=6∘A = 6^{\circ}, the angle of incidence at the second face is r2=A−r1=6∘−0∘=6∘r_2 = A - r_1 = 6^{\circ} - 0^{\circ} = 6^{\circ}.
  3. Snell's Law for Emergence: Applying Snell's law at the second face (from the prism to air): μsin⁡(r2)=1⋅sin⁡(e)\mu \sin(r_2) = 1 \cdot \sin(e). Since the angles are small, we can use the small-angle approximation sin⁡(θ)≈θ\sin(\theta) \approx \theta.
  4. Calculation: Substituting the given values, e≈μ×r2=1.5×6∘=9∘e \approx \mu \times r_2 = 1.5 \times 6^{\circ} = 9^{\circ}. Therefore, the angle of emergence is 9∘9^{\circ}.

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 INSTRUMENTShorizontalincidentrightangledrefractivematerial

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