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Zorluk: OrtaRefraction of Light, Total Internal Reflection, and Prisms

A light ray travels near the boundary between medium 1 (refractive index 1.601.60) and medium 2 (refractive index 1.201.20). For total internal reflection to occur at this boundary, in which direction must the light travel, and what is the sine of the critical angle (sin⁡C\sin C)?

  1. From medium 1 to medium 2, with sin⁡C=0.75\sin C = 0.75Cevap
  2. B
    From medium 2 to medium 1, with sin⁡C=0.75\sin C = 0.75
  3. C
    From medium 1 to medium 2, with sin⁡C=1.33\sin C = 1.33
  4. D
    From medium 2 to medium 1, with sin⁡C=1.33\sin C = 1.33

Cevap

Light must travel from medium 1 to medium 2, with sin⁡C=0.75\sin C = 0.75
Total internal reflection can only take place when light travels from an optically denser medium to an optically rarer medium. Here, medium 1 has a higher refractive index (1.601.60) than medium 2 (1.201.20), so light must travel from medium 1 to medium 2. The critical angle relationship n1sin⁡C=n2sin⁡90∘n_1 \sin C = n_2 \sin 90^\circ yields sin⁡C=n2n1=1.201.60=0.75\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.60} = 0.75.

Adım Adım Çözüm

1
Determine the required direction of light propagation for total internal reflection.
Light must travel from medium 1 (n1=1.60n_1 = 1.60) to medium 2 (n2=1.20n_2 = 1.20), which is from an optically denser medium to a rarer medium.
Total internal reflection occurs only when light moves toward a medium of lower refractive index so that the angle of refraction can reach 90∘90^\circ.
2
Calculate the sine of the critical angle using Snell's law at the critical condition.
sin⁡C=n2n1=1.201.60=0.75\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.60} = 0.75.
Applying Snell's law n1sin⁡C=n2sin⁡90∘n_1 \sin C = n_2 \sin 90^\circ gives sin⁡C=n2n1\sin C = \frac{n_2}{n_1} since sin⁡90∘=1\sin 90^\circ = 1.

Anahtar Kavram

Conditions for Total Internal Reflection and Critical Angle Calculation
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Konular

Refraction of Light, Total Internal Reflection, and Prisms

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