Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows a crystalline slab (refractive index 1.665) with a rectangular cross section. A ray of light strikes the slab at an incident angle of 1 = 37.0°, enters the slab, and travels to point P. This slab is surrounded by a fluid with a refractive index n. What is the maximum value of n such that total internal reflection occurs at point P?

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-05-07T04:13:21+02:00

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle [tex]\theta _c = sin^{-1} (\frac{n}{n_{slab}} )[/tex]

[tex]sin \theta _ c =\frac{n}{n_{slab}}[/tex]

According to Snell's law of refraction:

[tex]n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)[/tex]

At point P ; [tex]90 - \theta _2 \leq \theta _c[/tex]

[tex]\theta _2 = 90 - \theta _c[/tex]

Therefore:

[tex]n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}[/tex]

Then maximum value of refractive index n of the fluid is:

[tex]n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }[/tex]

[tex]n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }[/tex]

n = 1.4266