Math Problem Statement

Determine the critical angle for quartz, using the Index of Refraction Table in Question 3.

[3 marks]

The critical angle for sapphire is 34.4°. For each angle of incidence, determine if it would result in total internal reflection in a sapphire by writing Yes/No. [4 marks]

a) 23.7° _____________________ c) 53.4° _____________________ b) 34.7° _____________________ d) 31.5° _____________________

  1. A diver shines a flashlight toward the air-water boundary at an angle of 35° to the normal. At what angle does the flashlight beam leave the water? Assume that the index of refraction for the water is 1.33. [3 marks]

Solution

Question 1: Determine the critical angle for quartz

To determine the critical angle (θc\theta_c) for quartz, we use Snell's Law: n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2

At the critical angle, θ2\theta_2 is 90° (the angle of refraction), so: n1sinθc=n2sin90n_1 \sin \theta_c = n_2 \sin 90^\circ n1sinθc=n2n_1 \sin \theta_c = n_2 sinθc=n2n1\sin \theta_c = \frac{n_2}{n_1} θc=arcsin(n2n1)\theta_c = \arcsin \left( \frac{n_2}{n_1} \right)

For total internal reflection to occur, light must travel from a medium with a higher refractive index (n1n_1) to one with a lower refractive index (n2n_2). Assuming n1n_1 is the refractive index of quartz and n2n_2 is the refractive index of air (approximately 1):

Given:

  • n1n_1 for quartz ≈ 1.54 (common value for quartz)

θc=arcsin(11.54)\theta_c = \arcsin \left( \frac{1}{1.54} \right) θc=arcsin(0.6494)\theta_c = \arcsin (0.6494) θc40.49\theta_c \approx 40.49^\circ

Question 2: Total Internal Reflection in Sapphire

Given the critical angle for sapphire is 34.4°, we need to determine if the following angles of incidence result in total internal reflection (TIR) in sapphire. TIR occurs when the angle of incidence is greater than the critical angle.

a) 23.7°
23.7<34.423.7^\circ < 34.4^\circ No

b) 34.7°
34.7>34.434.7^\circ > 34.4^\circ Yes

c) 53.4°
53.4>34.453.4^\circ > 34.4^\circ Yes

d) 31.5°
31.5<34.431.5^\circ < 34.4^\circ No

Question 3: Refraction of Light from Water to Air

Using Snell's Law to determine the angle at which the flashlight beam leaves the water:

Given:

  • n1n_1 (water) = 1.33
  • n2n_2 (air) ≈ 1
  • Angle of incidence in water (θ1\theta_1) = 35°

n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2 1.33sin35=1sinθ21.33 \sin 35^\circ = 1 \sin \theta_2 sinθ2=1.33sin35\sin \theta_2 = 1.33 \sin 35^\circ sinθ2=1.33×0.5736\sin \theta_2 = 1.33 \times 0.5736 sinθ2=0.7629\sin \theta_2 = 0.7629

Since sinθ2\sin \theta_2 must be less than or equal to 1, there's an error. We need to reassess our calculation as follows:

sinθ2=1.33sin351\sin \theta_2 = \frac{1.33 \sin 35^\circ}{1} sinθ2=1.33×0.57361\sin \theta_2 = \frac{1.33 \times 0.5736}{1} sinθ2=0.7629\sin \theta_2 = 0.7629

θ2=arcsin(0.7629)\theta_2 = \arcsin (0.7629) θ249.8\theta_2 \approx 49.8^\circ

Thus, the angle at which the flashlight beam leaves the water is approximately 49.8°.


Do you want more details or have any questions?

Here are some related questions you might be interested in:

  1. What is the general principle behind Snell's Law?
  2. How do you derive the formula for the critical angle?
  3. What factors affect the critical angle of a material?
  4. How does total internal reflection apply to optical fibers?
  5. What are some practical applications of total internal reflection?
  6. How does the index of refraction vary with wavelength?
  7. Can the critical angle change if the surrounding medium changes?
  8. How does refraction differ from reflection?

Tip: The critical angle is a unique property of the material and the medium into which the light is transitioning. It only occurs when light moves from a medium with a higher index of refraction to one with a lower index of refraction.

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Math Problem Analysis

Mathematical Concepts

Optics
Snell's Law
Index of Refraction

Formulas

Snell's Law: n₁ sin θ₁ = n₂ sin θ₂

Theorems

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Suitable Grade Level

University