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

A ray of light traveling inside a glass prism of refractive index 1.601.60 strikes the boundary with an adjacent oil medium of refractive index 1.201.20. What is the sine of the critical angle for total internal reflection at this glass-oil interface?

  1. 0.750.75Cevap
  2. B
    1.331.33
  3. C
    0.6250.625
  4. D
    0.800.80

Cevap

The sine of the critical angle at the glass-oil interface is 0.750.75.
For light traveling from a medium of refractive index n1n_1 into a medium of refractive index n2n_2 (where n1>n2n_1 > n_2), the critical angle CC satisfies sinC=n2n1\sin C = \frac{n_2}{n_1}. Substituting n1=1.60n_1 = 1.60 and n2=1.20n_2 = 1.20 gives sinC=1.201.60=0.75\sin C = \frac{1.20}{1.60} = 0.75.

Adım Adım Çözüm

1
Identify the refractive indices of the two media.
Denser medium (glass): n1=1.60n_1 = 1.60; less dense medium (oil): n2=1.20n_2 = 1.20.
Total internal reflection occurs when light originates in the optically denser medium and strikes the boundary with a less dense medium.
2
Apply Snell's law at the critical angle CC.
n1sinC=n2sin90    sinC=n2n1n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}.
At the critical angle of incidence, the angle of refraction in the second medium is 9090^\circ, so sin90=1\sin 90^\circ = 1.
3
Substitute the given numerical values to compute sinC\sin C.
sinC=1.201.60=34=0.75\sin C = \frac{1.20}{1.60} = \frac{3}{4} = 0.75.
Dividing the refractive index of the oil by that of the glass gives the exact sine of the critical angle.

Anahtar Kavram

Critical Angle between Two Media
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