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3 years ago

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A light ray traveling through medium 1, index of refraction n1n1n_1 = 1.75, reaches the interface between medium 1 and medium 2,

index of refraction n2n2 = 1.24. Part A At what minimum angle with respect to the normal must the ray be incident on the interface in order to be totally internally reflected?

1 answer:

horrorfan [7]3 years ago

7 0

Answer:

θ₁ = 35.32°

Explanation:

given,

refractive index of medium 1 = n₁ = 1.75

refractive index of medium 2 = n₂ = 1.24

condition to describe the refracted angle

$\theta_{refracted} + \theta_{reflected}=90^0$

$\theta_2 = 90^0-\theta_1$ ...(1)

Using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

θ₁ , θ₂ is the angle of incidence and refractive index

n₁. n₂ is the refractive index medium 1 and medium 2

1.75 x sin θ₁ = 1.24 x sin θ₂

From equation (1)

1.75 x sin θ₁ = 1.24 x sin (90-θ₁)

1.75 sin θ₁ = 1.24 cos θ₁

tan θ₁ = 0.708

θ₁ = 35.32°

Hence, angle of incidence is equal to θ₁ = 35.32°

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