Two transparent prisms A and B of different refractive indices are placed in contact to produce a rectangular block. Figure 12 shows the path of a ray, incident normally on A, refracting as it crosses the boundary between the prisms. **Figure 12** A refractive index: 1.62 B refractive index: 1.35 not to scale P air **06.1** Explain how the path of the ray shows that the refractive index of A is greater than the refractive index of B. **06.2** Show that the angle of refraction of the ray in B is about 60°. **06.3** Draw, on Figure 12, the path of the ray immediately after it reaches P. Justify your answer with calculations.

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Two transparent prisms A and B of different refractive indices are placed in contact to produce a rectangular block - AQA - A-Level Physics - Question 6 - 2022 - Paper 1

Question 6

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Two transparent prisms A and B of different refractive indices are placed in contact to produce a rectangular block. Figure 12 shows the path of a ray, incident nor... show full transcript

Worked Solution & Example Answer:Two transparent prisms A and B of different refractive indices are placed in contact to produce a rectangular block - AQA - A-Level Physics - Question 6 - 2022 - Paper 1

Step 1

06.1 Explain how the path of the ray shows that the refractive index of A is greater than the refractive index of B.

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Answer

The path of the ray indicates that as it travels from air into prism A, the ray bends towards the normal due to a higher refractive index (1.62) compared to air. When the ray reaches prism B, it bends away from the normal, indicating a lower refractive index (1.35). This behavior aligns with Snell's law, where a greater refractive index causes light to bend towards the normal.

Step 2

06.2 Show that the angle of refraction of the ray in B is about 60°.

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Answer

To find the angle of refraction in prism B, we apply Snell's Law:

$n_1 imes ext{sin}( heta_1) = n_2 imes ext{sin}( heta_2)$

Here,

$n_1 = 1.62$ (for prism A)
$heta_1 = 43°$ (the angle of incidence)
$n_2 = 1.35$ (for prism B)

Setting the values in the equation: $1.62 imes ext{sin}(43°) = 1.35 imes ext{sin}( heta_2)$

After calculating, we find: $heta_2 = ext{about } 60°$

Step 3

06.3 Draw, on Figure 12, the path of the ray immediately after it reaches P. Justify your answer with calculations.

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Answer

After the ray reaches point P, it again encounters the boundary going from prism B to air. We can continue using Snell's Law: Setting:

$n_2 = 1.35$ (for prism B)
$n_3 = 1.00$ (for air)
$heta_2 = 60°$ (angle of refraction from previous calculation)

Using Snell's Law again: $1.35 imes ext{sin}(60°) = 1.00 imes ext{sin}( heta_3)$

Calculating for $heta_3$ gives: $heta_3 = 75°$

Thus, the ray exits into the air at approximately 75° with the normal line.

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Two transparent prisms A and B of different refractive indices are placed in contact to produce a rectangular block - AQA - A-Level Physics - Question 6 - 2022 - Paper 1

Worked Solution & Example Answer:Two transparent prisms A and B of different refractive indices are placed in contact to produce a rectangular block - AQA - A-Level Physics - Question 6 - 2022 - Paper 1

06.1 Explain how the path of the ray shows that the refractive index of A is greater than the refractive index of B.

06.2 Show that the angle of refraction of the ray in B is about 60°.

06.3 Draw, on Figure 12, the path of the ray immediately after it reaches P. Justify your answer with calculations.

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