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° _____________________

6. 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]**

### Question 1: Determine the critical angle for quartz

To determine the critical angle (\(\theta_c\)) for quartz, we use Snell's Law:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]

At the critical angle, \(\theta_2\) is 90° (the angle of refraction), so:
\[ n_1 \sin \theta_c = n_2 \sin 90^\circ \]
\[ n_1 \sin \theta_c = n_2 \]
\[ \sin \theta_c = \frac{n_2}{n_1} \]
\[ \theta_c = \arcsin \left( \frac{n_2}{n_1} \right) \]

For total internal reflection to occur, light must travel from a medium with a higher refractive index (\(n_1\)) to one with a lower refractive index (\(n_2\)). Assuming \( n_1 \) is the refractive index of quartz and \( n_2 \) is the refractive index of air (approximately 1):

Given:
- \( n_1 \) for quartz ≈ 1.54 (common value for quartz)

\[ \theta_c = \arcsin \left( \frac{1}{1.54} \right) \]
\[ \theta_c = \arcsin (0.6494) \]
\[ \theta_c \approx 40.49^\circ \]

### 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°  
\[ 23.7^\circ < 34.4^\circ \]
No

b) 34.7°  
\[ 34.7^\circ > 34.4^\circ \]
Yes

c) 53.4°  
\[ 53.4^\circ > 34.4^\circ \]
Yes

d) 31.5°  
\[ 31.5^\circ < 34.4^\circ \]
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:
- \( n_1 \) (water) = 1.33
- \( n_2 \) (air) ≈ 1
- Angle of incidence in water (\( \theta_1 \)) = 35°

\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
\[ 1.33 \sin 35^\circ = 1 \sin \theta_2 \]
\[ \sin \theta_2 = 1.33 \sin 35^\circ \]
\[ \sin \theta_2 = 1.33 \times 0.5736 \]
\[ \sin \theta_2 = 0.7629 \]

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

\[ \sin \theta_2 = \frac{1.33 \sin 35^\circ}{1} \]
\[ \sin \theta_2 = \frac{1.33 \times 0.5736}{1} \]
\[ \sin \theta_2 = 0.7629 \]

\[ \theta_2 = \arcsin (0.7629) \]
\[ \theta_2 \approx 49.8^\circ \]

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

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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.

Explore the determination of the critical angle for quartz using Snell's Law and the assessment of total internal reflection in sapphire based on given angles of incidence. Learn about optics principles with detailed explanations and calculations.

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