Let's go through each question step by step.

1. A ray of light passes from a liquid to air. The angle of incidence and the angle of refraction are 35° and 53° respectively. Find the refractive index of the liquid.

To solve this, we use Snell's Law:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

Where:

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

Rearrange the equation to solve for $n_1$:

$n_1 = \frac{n_2 \sin(\theta_2)}{\sin(\theta_1)} = \frac{1 \cdot \sin(53^\circ)}{\sin(35^\circ)}$

Using sine values: $n_1 = \frac{0.7986}{0.5736} \approx 1.39$

Answer: 1.39 (which you wrote down correctly)

2. A light ray travels from air to glass as shown (angles 25° and 50°). What is the refractive index of the glass?

Using Snell's Law again: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$ Where:

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

Rearrange to find $n_2$: $n_2 = \frac{n_1 \sin(\theta_1)}{\sin(\theta_2)} = \frac{1 \cdot \sin(25^\circ)}{\sin(50^\circ)}$

Using sine values: $n_2 = \frac{0.4226}{0.7660} \approx 0.55$

Answer: $n_2 \approx 0.55$

However, you wrote 1.4 as the answer. Did you use different assumptions or angles? Let me know if you'd like to discuss this further.

3. A light ray travels from air to glass (angle of refraction given as 35°). If the refractive index of the glass is 1.5, what is the angle of incidence?

Again using Snell's Law: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$ Where:

• $n_1 = 1$ (air),
• $n_2 = 1.5$ (glass),
• $\theta_2 = 35^\circ$,
• $\theta_1$ is the angle of incidence (unknown).

Rearrange to solve for $\theta_1$: $\sin(\theta_1) = \frac{n_2 \sin(\theta_2)}{n_1} = \frac{1.5 \cdot \sin(35^\circ)}{1}$

$\sin(\theta_1) = 1.5 \cdot 0.5736 = 0.8604$

Now, find $\theta_1$ by taking the inverse sine: $\theta_1 = \sin^{-1}(0.8604) \approx 59.5^\circ$

Answer: 59.5°

4. Light ray XY passes from glass into air and back into glass. Which path (1, 2, 3, or 4) correctly indicates the subsequent path of the light ray?

In this case, the light is passing from a denser medium (glass) to a less dense medium (air), then back into a denser medium. The ray will bend according to the laws of refraction. Based on the diagram, Path 2 is the correct direction, as it follows the expected behavior of light bending toward the normal when entering glass from air.

Answer: 2

Would you like more details on any specific question? Here are five additional related questions for practice:

1. What is the relationship between the refractive index and the speed of light in a medium?
2. How would the refractive index change if the light ray traveled from air to water instead of air to glass?
3. What happens to a light ray when it passes through a medium with a refractive index of less than 1?
4. How would you calculate the critical angle for total internal reflection when light passes from glass to air?
5. If the angles were reversed in the second question (50° in air, 25° in glass), how would the refractive index change?

Tip: Always remember to check whether the light is moving into a denser or less dense medium, as it affects how much the light will bend.