Question:
In figure, the optical fiber is \(l\) = 2m long and has a diameter of d = 20 \(\mu\)m. 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)
In figure, the optical fiber is \(l\) = 2m long and has a diameter of d = 20 \(\mu\)m. 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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Use Snell's law at the entry face to get \(\theta_2\), then each reflection advances the ray by \(d/\tan\theta_2\) along the fibre; divide the length by this.
Updated On: Nov 5, 2026
- 57000
- 55000
- 66000
- 45000
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The Correct Option is A
Solution and Explanation
Snell's law at the end face: \(\sin\theta_1 = n\sin\theta_2\), so \(\sin\theta_2 = \dfrac{0.64}{1.31} = 0.4885\).
\(\cos\theta_2 = \sqrt{1-0.2387} = 0.8725\), hence \(\tan\theta_2 = \dfrac{0.4885}{0.8725} = 0.56\) (\(\theta_2\) is the angle of the ray with the fibre axis).
Between two successive reflections the ray advances along the axis by \(x = \dfrac{d}{\tan\theta_2} = \dfrac{20\times10^{-6}}{0.56} \approx 3.57\times10^{-5}\) m.
Number of reflections \(N = \dfrac{l}{x} = \dfrac{2}{3.57\times10^{-5}} \approx 5.6\times10^{4}\).
The closest option is 57000.
Answer: (1) 57000
\(\cos\theta_2 = \sqrt{1-0.2387} = 0.8725\), hence \(\tan\theta_2 = \dfrac{0.4885}{0.8725} = 0.56\) (\(\theta_2\) is the angle of the ray with the fibre axis).
Between two successive reflections the ray advances along the axis by \(x = \dfrac{d}{\tan\theta_2} = \dfrac{20\times10^{-6}}{0.56} \approx 3.57\times10^{-5}\) m.
Number of reflections \(N = \dfrac{l}{x} = \dfrac{2}{3.57\times10^{-5}} \approx 5.6\times10^{4}\).
The closest option is 57000.
Answer: (1) 57000
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