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NEET PHYSICSMedium

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}.
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