Math Problem Statement

Calcite (CaCO3) is a crystal with abnormally large birefringence. The index of refraction for light with electric field parallel to the optical axis (called extraordinary waves or e-waves) is 1.4864. The index of refraction for light with electric field perpendicular to the optical axis (called ordinary waves or o-waves) is 1.6584. Part A Find the critical angle θce for e-waves in calcite. Express your answer in degrees to four significant figures. View Available Hint(s) for Part A Hint 1 for Part A. Snell's law Recall that Snell's law states that, when a beam of light is incident on an interface between a material with index of refraction n1 and a material with index of refraction n2, the relation between the angles (with respect to the normal to the interface) is given by the equation n1sin(θ1)=n2sin(θ2). Hint 2 for Part A. Definition of critical angle The critical angle is defined as the angle to the normal at which rays moving from higher index of refraction to lower index of refraction (for instance, leaving a block of calcite into air) would have angle of refraction equal to 90∘. Incident angles larger than the critical angle lead to total internal reflection of the ray. Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

θce =

Solution

To find the critical angle θce\theta_{ce} for e-waves in calcite, we’ll use Snell's law and the concept of the critical angle:

  1. Snell's Law: n1sin(θ1)=n2sin(θ2)n_1 \sin(\theta_1) = n_2 \sin(\theta_2)

  2. Critical Angle Condition: The critical angle θce\theta_{ce} is defined as the angle of incidence in calcite for which the angle of refraction in air (θ2\theta_2) is 90°.

Since the refractive index of air (n2n_2) is approximately 1.0001.000, and the refractive index for e-waves in calcite (n1n_1) is 1.48641.4864, we set θ2=90\theta_2 = 90^\circ and rearrange Snell's law as follows:

sin(θce)=n2n1\sin(\theta_{ce}) = \frac{n_2}{n_1}

  1. Calculate θce\theta_{ce}: sin(θce)=1.0001.4864\sin(\theta_{ce}) = \frac{1.000}{1.4864}

    Solving for θce\theta_{ce}:

    θce=arcsin(1.0001.4864)\theta_{ce} = \arcsin\left(\frac{1.000}{1.4864}\right)

Let's compute this.import math

Given values

n_air = 1.000 # Refractive index of air n_calcite_e_wave = 1.4864 # Refractive index for e-waves in calcite

Calculating the critical angle in degrees

theta_ce_radians = math.asin(n_air / n_calcite_e_wave) theta_ce_degrees = math.degrees(theta_ce_radians) theta_ce_degrees42.2809388169988442.28093881699884successThe critical angle θce\theta_{ce} for e-waves in calcite is approximately 42.2842.28^\circ (to four significant figures).

Would you like further details on this calculation, or do you have any other questions?

Here are five related questions for deeper exploration:

  1. How would the critical angle change if the refractive index for e-waves were slightly higher or lower?
  2. How does birefringence in calcite affect the behavior of o-waves compared to e-waves?
  3. What is the critical angle for o-waves in calcite, given the refractive index for o-waves?
  4. How does total internal reflection depend on the critical angle in materials with high birefringence?
  5. In what practical applications is the birefringence of calcite used, especially concerning polarized light?

Tip: For highly birefringent materials like calcite, critical angles will differ for ordinary and extraordinary waves, making them useful in polarizing optics.

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Math Problem Analysis

Mathematical Concepts

Optics
Refraction
Critical Angle
Snell's Law

Formulas

Snell's Law: n1 * sin(θ1) = n2 * sin(θ2)
Critical Angle Formula: sin(θc) = n2 / n1 for θ2 = 90°

Theorems

Snell's Law
Critical Angle Theorem

Suitable Grade Level

College/University Level Physics