Question

A layer of methyl alcohol, with index of refraction $1.329,$ rests on a block of ice, with index of refraction $1.310 .$ A ray of light passes through the methyl alcohol at an angle of $\varphi_{1}=61.07^{\circ}$ relative to the alcohol-ice boundary. What is the angle $\varphi_{2}$ relative to the boundary at which the ray passes through the ice?

Name: Problem 69, Chapter 32: University Physics with Modern Physics
Uploaded: 2021-05-20T09:19:39-08:00
Duration: 2 min 19 s
Channel: Narayan Hari
Description: A layer of methyl alcohol, with index of refraction 1.329, rests on a block of ice, with index of refraction 1.310 . A ray of light passes through the methyl alcohol at an angle of φ1=61.07^∘ relative to the alcohol-ice boundary. What is the angle φ2 relative to the boundary at which the ray passes through the ice?

   A layer of methyl alcohol, with index of refraction $1.329,$ rests on a block of ice, with index of refraction $1.310 .$ A ray of light passes through the methyl alcohol at an angle of $\varphi_{1}=61.07^{\circ}$ relative to the alcohol-ice boundary. What is the angle $\varphi_{2}$ relative to the boundary at which the ray passes through the ice?

University Physics with Modern Physics

Wolfgang Bauer, Gary… 2nd Edition

Chapter 32, Problem 69

Instant Answer

Solved by Expert Narayan Hari

05/20/2021

Step 1

This can be calculated as $\theta_{1} = 90^{\circ} - \varphi_{1}$, where $\varphi_{1}$ is the angle of the light ray relative to the alcohol-ice boundary. Substituting the given value of $\varphi_{1}$, we get \[\theta_{1} = 90^{\circ} - 61.07^{\circ} = Show more…

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A layer of methyl alcohol, with index of refraction $1.329,$ rests on a block of ice, with index of refraction $1.310 .$ A ray of light passes through the methyl alcohol at an angle of $\varphi_{1}=61.07^{\circ}$ relative to the alcohol-ice boundary. What is the angle $\varphi_{2}$ relative to the boundary at which the ray passes through the ice?

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Key Concepts

Index of Refraction

This is a fundamental property of a material that quantifies how much light slows down when it enters the material relative to its speed in a vacuum. It also determines how much light bends, or refracts, when transitioning between materials with different indexes. In problems involving light propagation through different media, knowing each material's index of refraction is essential for predicting the direction of the transmitted beam.

Snell's Law

Snell's Law is the principle that relates the angles of incidence and refraction to the indexes of refraction of the two media. It is expressed as n? sin(??) = n? sin(??), where ?? and ?? are the angles measured relative to the normal of the interface. In cases where angles are given relative to the boundary (rather than the normal), appropriate geometric adjustments are made. This law is crucial for determining how a light ray will bend when traveling from one medium to another.

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