Question

Step 1 The angles of incidence ?1i from air to glass and ?2i from glass to air, and the angles of refraction ?1r in the glass and ?2r in the air as the light ray exits the prism are shown in the diagram. (a) The angle of incidence at the first air-glass surface is ?1i = 40.9°. We know that the index of refraction of air is n_air = 1.0 and the index of refraction of this glass n_glass = 1.5. Snell's law gives the angle of refraction at the first air-glass surface, where sin ?1i = sin 40.9° = 0.6547. ?1r = sin?¹( (n_air sin ?1i) / n_glass ) = sin?¹( 1.0(0.6547) / 1.5 ) = 25.87° ? 25.9°. Step 2 From the diagram, we observe that ?1r + ? = 90°, so we have ? = 90° ? ?1r. Since we know that the sum of the interior angles of a triangle is 180°, ? + ? + 60° = 180°. Therefore, the angle the refracted ray makes with the second surface is ? = 120° ? ? = 120° ? (90° ? ?1r) = 30° + ?1r = 30° + 25.9° = 55.9°. The angle of incidence at the second surface is then ?2i = 90° ? ? = 90° ? 55.9° = 34.1°.

          Step 1
The angles of incidence ?1i from air to glass and ?2i from glass to air, and the angles of refraction ?1r in the glass and ?2r in the air as the light ray exits the prism are shown in the diagram.
(a) The angle of incidence at the first air-glass surface is ?1i = 40.9°. We know that the index of refraction of air is n_air = 1.0 and the index of refraction of this glass n_glass = 1.5. Snell's law gives the angle of refraction at the first air-glass surface, where sin ?1i = sin 40.9° = 0.6547.
?1r = sin?¹( (n_air sin ?1i) / n_glass ) = sin?¹( 1.0(0.6547) / 1.5 ) = 25.87° ? 25.9°.

Step 2
From the diagram, we observe that ?1r + ? = 90°, so we have ? = 90° ? ?1r.
Since we know that the sum of the interior angles of a triangle is 180°, ? + ? + 60° = 180°.
Therefore, the angle the refracted ray makes with the second surface is
? = 120° ? ? = 120° ? (90° ? ?1r) = 30° + ?1r = 30° + 25.9° = 55.9°.
The angle of incidence at the second surface is then
?2i = 90° ? ? = 90° ? 55.9° = 34.1°.

$Step 1 The angles of incidence ?1i from air to glass and ?2i from glass to air, and the angles of refraction ?1r in the glass and ?2r in the air as the light ray exits the prism are shown in the diagram. (a) The angle of incidence at the first air-glass surface is ?1i = 40.9°. We know that the index of refraction of air is nair = 1.0 and the index of refraction of this glass nglass = 1.5. Snell's law gives the angle of refraction at the first air-glass surface, where sin ?1i = sin 40.9° = 0.6547. ?1r = sin?¹( (nair sin ?1i) / nglass ) = sin?¹( 1.0(0.6547) / 1.5 ) = 25.87° ? 25.9°. Step 2 From the diagram, we observe that ?1r + ? = 90°, so we have ? = 90° ? ?1r. Since we know that the sum of the interior angles of a triangle is 180°, ? + ? + 60° = 180°. Therefore, the angle the refracted ray makes with the second surface is ? = 120° ? ? = 120° ? (90° ? ?1r) = 30° + ?1r = 30° + 25.9° = 55.9°. The angle of incidence at the second surface is then ?2i = 90° ? ? = 90° ? 55.9° = 34.1°.$

Added by David C.

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