9.
Light is incident from a medium with refractive index n = 1.72 onto vacuum. The
smallest angle of incidence (measured with respect to the normal) for which light is not
transmitted into vacuum is (approximately):
a. 30°
b. 45°
c. 60°
d. 90°
The critical angle for total internal reflection is approximately 36°, so the smallest angle of incidence where light is not transmitted into vacuum is just above this value, closest to option (b) 45° among the choices.
Problem Solution
Light travels from a denser medium (n₁ = 1.72) to vacuum (n₂ = 1.00). Total internal reflection (TIR) occurs when the angle of incidence θ_i exceeds the critical angle θ_c, defined by Snell’s law at the boundary where the refraction angle is 90°: n₁ sin θ_c = n₂ sin 90°. Thus, sin θ_c = n₂ / n₁ = 1 / 1.72 ≈ 0.5814, so θ_c = sin⁻¹(0.5814) ≈ 35.55°. Any θ_i ≥ 35.55° prevents transmission into vacuum.
Option Analysis
- a. 30°: Below θ_c (35.55°), so light refracts into vacuum with partial transmission. Incorrect.
- b. 45°: Exceeds θ_c, causing TIR with no transmission. Closest match to the smallest such angle among options. Correct.
- c. 60°: Also exceeds θ_c, but larger than necessary; not the smallest. Incorrect.
- d. 90°: Grazing incidence causes TIR, but far from smallest angle. Incorrect.
TIR Formula and Calculation
The critical angle formula for TIR is θ_c = sin⁻¹(n₂/n₁), where n₁ > n₂. Here, θ_c ≈ 35.55°, so transmission stops for θ_i > 35.55°. Real-world applications include optical fibers, where high-n cores (like n=1.72) ensure TIR for signal transmission.
Option Comparison Table
| Option | Angle | Outcome (vs θ_c=35.55°) | Result |
|---|---|---|---|
| a | 30° | Below | Transmits |
| b | 45° | Above (smallest match) | TIR |
| c | 60° | Above | TIR |
| d | 90° | Above | TIR |
This table highlights why 45° is the answer, ideal for CSIR NET physics preparation on refraction and TIR.
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