In figure, the optical fiber is l = 2m long and has a diameter of d = 20 μm. If a ray of light is incident on one end of the fiber at angle θ₁ = 40°, the number of reflection it makes before emerging from the other end is close to: (refractive index of fibre is 1.31 and sin 40° = 0.64) (a) 57000 (b) 55000 (c) 66000 (d) 45000

Answer: (a). rdd / tan r Refraction at entry: sin r = (sin 40°)/(1.31) = (0.64)/(1.31) ≈ 0.49, so r ≈ 30° (angle of the ray with the fibre axis). Between two reflections the ray moves (d)/(tan r) along the axis. N = (ltan r)/(d) = 2 × tan 30°20 × 10⁻⁶ = 2 × 0.5772 × 10⁻⁵ ≈ 5.7 × 10⁴ So the answer is (a) 57000.

In figure, the optical fiber is l = 2m long and has a diameter of d =… | JEE Main 2019 (8 Apr, Shift 1)

Q45JEE Main 2019 · 8 APR · SHIFT 1Difficult

Total Internal Reflection

In figure, the optical fiber is $l$ = 2m long and has a diameter of d = 20 μm. If a ray of light is incident on one end of the fiber at angle $\theta_1$ = 40°, the number of reflection it makes before emerging from the other end is close to: (refractive index of fibre is 1.31 and sin 40° = 0.64)

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Answer: (a)

Refraction at entry: $\sin r = \frac{\sin 40^\circ}{1.31} = \frac{0.64}{1.31} \approx 0.49$ , so $r \approx 30^\circ$ (angle of the ray with the fibre axis).
Between two reflections the ray moves $\frac{d}{\tan r}$ along the axis.
$N = \frac{l\tan r}{d} = \frac{2 \times \tan 30^\circ}{20 \times 10^{-6}} = \frac{2 \times 0.577}{2 \times 10^{-5}} \approx 5.7 \times 10^4$
So the answer is (a) 57000.

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