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The figure below shows the path of a light beam through several slabs with different indices of refraction. (n4 = 1.12) 0₂ n = 1.60 n=1.40 n = 1.20 (a) If 0₁ = 34.0°, what is the angle 82 of the emerging beam? (b) What must the incident angle, be to have total internal reflection at the surface between the medium with n = 1.20 and the medium with n4 = 1.12?

The figure below shows the path of a light beam through several slabs with different indices of refraction. (n4 = 1.12) 0₂ n = 1.60 n=1.40 n = 1.20 (a) If 0₁ = 34.0°, what is the angle 82 of the emerging beam? (b) What must the incident angle, be to have total internal reflection at the surface between the medium with n = 1.20 and the medium with n4 = 1.12?

College Physics
College Physics
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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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The image illustrates the path of a light beam passing through several slabs with different indices of refraction. The indices are labeled as follows: - The first medium has an index of refraction \( n = 1.60 \). - The second medium has an index of refraction \( n = 1.40 \). - The third medium has an index of refraction \( n = 1.20 \). - The final medium, where the beam is emerging, has an index of refraction \( n_4 = 1.12 \). There are several arrows depicting the direction of the light beam as it refracts through each medium. The angles relevant to the light's path are denoted as \( \theta_1 \), the angle of incidence in the first medium, and \( \theta_2 \), the angle of the emerging beam in the last medium. The questions posed are: (a) If \( \theta_1 = 34.0^\circ \), what is the angle \( \theta_2 \) of the emerging beam? An input box is provided for the answer. (b) What must the incident angle \( \theta_1 \) be to have total internal reflection at the surface between the medium with \( n = 1.20 \) and the medium with \( n_4 = 1.12 \)? An input box is provided for the answer.
Transcribed Image Text:The image illustrates the path of a light beam passing through several slabs with different indices of refraction. The indices are labeled as follows: - The first medium has an index of refraction \( n = 1.60 \). - The second medium has an index of refraction \( n = 1.40 \). - The third medium has an index of refraction \( n = 1.20 \). - The final medium, where the beam is emerging, has an index of refraction \( n_4 = 1.12 \). There are several arrows depicting the direction of the light beam as it refracts through each medium. The angles relevant to the light's path are denoted as \( \theta_1 \), the angle of incidence in the first medium, and \( \theta_2 \), the angle of the emerging beam in the last medium. The questions posed are: (a) If \( \theta_1 = 34.0^\circ \), what is the angle \( \theta_2 \) of the emerging beam? An input box is provided for the answer. (b) What must the incident angle \( \theta_1 \) be to have total internal reflection at the surface between the medium with \( n = 1.20 \) and the medium with \( n_4 = 1.12 \)? An input box is provided for the answer.
Expert Solution
Step 1: Given Informations

Physics homework question answer, step 1, image 1

In medium 1:

The angle of incidence on the bottom surface is theta subscript 1.

refractive index is n equals n subscript 1 equals 1.60

In medium 2:

The angle of refraction at the top surface is theta subscript text a end text end subscript.

refractive index is n equals n subscript 2 equals 1.40

The angle of incidence on the bottom surface is theta subscript b

Here, theta subscript text a end text end subscript equals theta subscript text b end text end subscript (alternate angles)

In medium 3:

The angle of refraction at the top surface is theta subscript text c end text end subscript.

refractive index is n equals n subscript 3 equals 1.40

The angle of incidence on the bottom surface is theta subscript text d end text end subscript.

Here, theta subscript text c end text end subscript equals theta subscript text d end text end subscript (alternate angles)

In medium 4:

The angle of refraction at the top surface is theta subscript 4.

Refractive index is n equals n subscript 4 equals 1.12




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