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NEET PHYSICSMedium

A light ray falls on a glass surface of refractive index 3\sqrt{3}, at an angle of 60∘60^{\circ}. The angle between the refracted and reflected rays would be:

A

120^{\circ}

B

30^{\circ}

C

60^{\circ}

D

90^{\circ}

Step-by-Step Solution

  1. Check for Brewster's Angle: The relationship between the polarizing angle (ipi_p) and refractive index (μ\mu) is given by Brewster's Law: tan⁡ip=μ\tan i_p = \mu.
  2. Verification: Given μ=3\mu = \sqrt{3} and angle of incidence i=60∘i = 60^{\circ}. tan⁡60∘=3\tan 60^{\circ} = \sqrt{3} Since the given angle of incidence satisfies the condition tan⁡i=μ\tan i = \mu, the incident angle is the Brewster's angle.
  3. Property: At Brewster's angle, the reflected ray and the refracted ray are perpendicular to each other.
  4. Alternative Calculation:
  • Angle of reflection r′=i=60∘r' = i = 60^{\circ}.
  • Angle of refraction rr from Snell's Law: 1⋅sin⁡60∘=3sin⁡r⇒32=3sin⁡r⇒sin⁡r=12⇒r=30∘1 \cdot \sin 60^{\circ} = \sqrt{3} \sin r \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin r \Rightarrow \sin r = \frac{1}{2} \Rightarrow r = 30^{\circ}.
  • Angle between rays =180∘−(r′+r)=180∘−(60∘+30∘)=90∘= 180^{\circ} - (r' + r) = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}.
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