Figure 4 shows a ray of monochromatic light incident at angle A from air onto the end of a straight optical fibre. This ray undergoes total internal reflection at the core-cladding boundary. A ray that enters the optical fibre at an angle greater than A will only be partially reflected at the core–cladding boundary. Table 2 shows some properties of the optical fibre. Calculate the speed of the light ray in the optical fibre. Calculate A, in degrees, for the optical fibre shown in Figure 4. A ray is incident on the optical fibre at angle A. The optical fibre is now bent, as shown in Figure 5. Draw, on Figure 5, what happens to the ray within the optical fibre. Explain your answer.

A-Level AQA Physics Refraction: Figure 4 shows a ray of monochro

Figure 4 shows a ray of monochromatic light incident at angle A from air onto the end of a straight optical fibre - AQA - A-Level Physics - Question 3 - 2021 - Paper 1

Question 3

Figure 4 shows a ray of monochromatic light incident at angle A from air onto the end of a straight optical fibre. This ray undergoes total internal reflection at t... show full transcript

Worked Solution & Example Answer:Figure 4 shows a ray of monochromatic light incident at angle A from air onto the end of a straight optical fibre - AQA - A-Level Physics - Question 3 - 2021 - Paper 1

Step 1

Calculate the speed of the light ray in the optical fibre.

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Answer

To find the speed of light in the optical fibre, we can use the formula:

$v = rac{c}{n}$

where:

$c = 3.0 imes 10^8 ext{ m/s}$ (speed of light in a vacuum)
$n = 1.47$ (refractive index of the core)

Calculating gives: $v = rac{3.0 imes 10^8 ext{ m/s}}{1.47} \approx 2.04 \times 10^8 ext{ m/s}$

Step 2

Calculate A, in degrees, for the optical fibre shown in Figure 4.

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Answer

To calculate the critical angle, we use Snell's law:

$ext{sin}(C) = rac{n_{2}}{n_{1}}$

where:

$n_{1} = 1.47$ (refractive index of the core)
$n_{2} = 1.41$ (refractive index of the cladding)

Thus: $ext{sin}(C) = rac{1.41}{1.47} \approx 0.957$

Now, taking the inverse sine gives: $C \approx 73.6^{ ext{o}}$

As angle A must be less than the critical angle: $A < 73.6^{ ext{o}}$

Step 3

Draw, on Figure 5, what happens to the ray within the optical fibre. Explain your answer.

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Answer

A diagram should be drawn to show the ray bending as it encounters the optical fibre boundary. The ray will partially reflect at the core-cladding boundary based on its angle of incidence.

If the angle of incidence is greater than the critical angle (73.6°), some light will be reflected back into the core while some light may refract into the cladding.
If the angle of incidence is less than the critical angle, the ray will continue in the core.

Overall, as the fibre bends, it may cause variations in how the light travels, likely resulting in some light being refracted out into the cladding.

Figure 4 shows a ray of monochromatic light incident at angle A from air onto the end of a straight optical fibre - AQA - A-Level Physics - Question 3 - 2021 - Paper 1

Worked Solution & Example Answer:Figure 4 shows a ray of monochromatic light incident at angle A from air onto the end of a straight optical fibre - AQA - A-Level Physics - Question 3 - 2021 - Paper 1

Calculate the speed of the light ray in the optical fibre.

Calculate A, in degrees, for the optical fibre shown in Figure 4.

Draw, on Figure 5, what happens to the ray within the optical fibre. Explain your answer.

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