The critical angle for light passing from glass (n=1.5) to air is approximately:
A. 42°
B. 90°
C. 30°
D. 60°
Answer: Option A
Solution (By JKSSB Mock Tests)
Critical angle θ_c for total internal reflection: sinθ_c = n₂/n₁, where n₁ > n₂. Here n₁=1.5 (glass), n₂=1 (air), so sinθ_c = 1/1.5 = 2/3 ≈ 0.6667 ⇒ θ_c = sin⁻¹(0.6667) ≈ 41.8° ≈ 42°. Memory aid: 'sinθ_c = 1/n for glass to air'. This calculation tests understanding of total internal reflection conditions, crucial for optics in competitive exams. Always ensure light travels from denser to rarer medium for TIR to be possible; otherwise, no critical angle exists.
Explanation:
Work W = F·d = Fd cosθ. cos0° = 1 gives maximum work. At 90°, cos90° = 0, work zero. For given magnitudes, maximum when force parallel to displacement. This is basic definition.
No comments yet. Be the first to start the discussion!