In figure, the optical fiber is l=2m long and has a diameter of d=20μm. If a ray of light is incident on one end of the fiber at angle θ1=40∘, the number of reflections it makes before emerging from the other end is close to:(Refractive index of fiber is 1.31 and sin⁡40∘=0.64)

Using Snell’s law of refraction,1×sin⁡40∘=1.31sin⁡θ⇒sin⁡θ=0.641.31=0.49≈0.5⇒θ=30∘∴Number of reflections=220×10−6×cot⁡θ=2×10620×3=57735≈57000

In figure, the optical fiber is l=2m long and has a diameter of d=20μm. If a ray of light is incident on one end of the fiber at angle θ1=40∘, the number of reflections it makes before emerging from the other end is close to:(Refractive index of fiber is 1.31 and sin⁡40∘=0.64)

Using Snell’s law of refraction,1×sin⁡40∘=1.31sin⁡θ⇒sin⁡θ=0.641.31=0.49≈0.5⇒θ=30∘∴Number of reflections=220×10−6×cot⁡θ=2×10620×3=57735≈57000

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In figure, the optical fiber is $l = 2 m$ long and has a diameter of $d = 20 μm$ . If a ray of light is incident on one end of the fiber at angle $θ_{1} = 40^{\circ}$ , the number of reflections it makes before emerging from the other end is close to:
(Refractive index of fiber is 1.31 and $\sin (40^{\circ}) = 0.64$ )

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45000

75000

66000

57000

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Detailed Solution

Using Snell’s law of refraction,

$1 \times \sin 40^{\circ} = 1 .31 \sin θ$

$\Rightarrow \sin θ = \frac{0 .64}{1 .31} = 0 .49 \approx 0 .5 \Rightarrow θ = 30^{\circ}$

$∴$ Number of reflections

$= \frac{2}{20 \times 10^{- 6} \times \cot θ}$

$= \frac{2 \times 10^{6}}{20 \times \sqrt{3}} = 57735 \approx 57000$

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In figure, the optical fiber is l=2m long and has a diameter of d=20μm. If a ray of light is incident on one end of the fiber at angle θ1=40∘, the number of reflections it makes before emerging from the other end is close to:(Refractive index of fiber is 1.31 and sin⁡40∘=0.64)

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