Let's go through the questions from your Physics Wave Motion I worksheet one by one.

1. A ray of light passes from a liquid to air.

• The angle of incidence and the angle of refraction are given as 35° and 53°, respectively. You are asked to find the refractive index of the liquid.

Solution Approach: Use Snell's Law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$

• $n_1$ = refractive index of the liquid (unknown)
• $n_2 = 1$ (refractive index of air)
• $\theta_1 = 35^\circ$ (angle of incidence)
• $\theta_2 = 53^\circ$ (angle of refraction)

Rearranging for $n_1$: $n_1 = \frac{n_2 \sin \theta_2}{\sin \theta_1} = \frac{1 \cdot \sin 53^\circ}{\sin 35^\circ}$ Calculate the values of $\sin 53^\circ$ and $\sin 35^\circ$ to find the refractive index.

2. A light ray travels from air to glass.

• The angle of incidence in air is 25° and the angle of refraction in glass is 50°.
• You are asked to find the refractive index of the glass.

Solution Approach: Again, use Snell's Law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$

• $n_1 = 1$ (refractive index of air)
• $\theta_1 = 25^\circ$ (angle of incidence in air)
• $\theta_2 = 50^\circ$ (angle of refraction in glass)
• $n_2$ = refractive index of glass (unknown)

Rearrange for $n_2$: $n_2 = \frac{n_1 \sin \theta_1}{\sin \theta_2} = \frac{1 \cdot \sin 25^\circ}{\sin 50^\circ}$

3. A light ray travels from air to glass.

• The refractive index of the glass is given as 1.5, and the angle of refraction is 55°.
• You are asked to find the angle of incidence in air.

Solution Approach: Use Snell's Law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$

• $n_1 = 1$ (refractive index of air)
• $n_2 = 1.5$ (refractive index of glass)
• $\theta_2 = 55^\circ$ (angle of refraction)
• $\theta_1$ = angle of incidence (unknown)

Rearrange for $\sin \theta_1$: $\sin \theta_1 = \frac{n_2 \sin \theta_2}{n_1} = \frac{1.5 \cdot \sin 55^\circ}{1}$ Then find $\theta_1$ by taking the inverse sine.

4. Diagram: Light ray passing through a triangular wedge of air and glass.

• You need to determine which of the numbered paths (1, 2, 3, or 4) correctly indicates the subsequent path of the light ray after it re-enters the glass.

Solution Approach: Apply Snell's Law and basic refraction principles at each interface (air to glass and back to air). Based on the angles and refractive indices, determine the correct path number.

Would you like detailed solutions for each problem or have any specific questions? Here are five related questions for practice:

1. What is the physical significance of the refractive index in optics?
2. How would the angles change if the light were traveling from a denser medium to a rarer medium (e.g., glass to air)?
3. How does total internal reflection occur, and under what conditions?
4. Can you derive Snell's Law from the wave theory of light?
5. How would you experimentally determine the refractive index of an unknown liquid?

Tip: Always remember that the refractive index is a ratio of the speed of light in vacuum to the speed in the medium. Hence, it is always greater than or equal to 1.