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kari74 [83]
3 years ago
9

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?
Physics
1 answer:
Vlada [557]3 years ago
3 0

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 \theta _c = sin^{-1} (\frac{n}{n_{slab}} )

sin \theta _ c =\frac{n}{n_{slab}}

According to Snell's law of refraction:

n sin \theta _1 = n_{slab}  \ sin \  (90- \theta_c)

At point P ; 90 - \theta _2  \leq \theta _c

\theta _2 = 90 - \theta _c

Therefore:

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

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

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

n = \frac{1.665}{\sqrt{1+ sin^2 \  37} }

n = 1.4266

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